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OXFORD INTERNATIONAL PUBLIC SCHOOL

CLASS-XI ( SCIENCE)

HOLIDAY HOME WORK

Sl.no SUBJECT DETAILS

1. English Prepare a fileon the ideal of your lkife with creative writing and beautiful

drawings.

2

Biology

Instruction—

1. Select one project for holiday homework.

2. Use filepages.(don’t use rough pages)

3. You can use this below url to get more instruction or answer

https://studyres.com/doc/1861347/investigatory-project-for-class-xii-

biology

4.important project number-5,6,7,12,13,14,15

5.after selecting the project ,send me message(your name,project name)before

starting the project.

INVESTIGATORY PROJECT FOR CLASS XII BIOLOGY

1. Project Report on Malnutrition

2. Biology Project Report on Components of Food

3. Biology Project Report on DNA Fingerprinting

4. Project Report on Pollution

5. Biology Project Report on ABO blood grouping in human beings

6. Biology Project Report on the dispersal of seeds by various agencies

7. Biology Project Report on mosquito species -major diseases caused by

it

8. Biology Project Report on Human diseases

9. Project Report on sleeping Habits in human beings

10. Biology Project Report on Manures and Chemical Fertilizers

11. Project Report on Useful Plants and Animals

12. Biology Project Report on Cancer

13. Biology Project Report on aids

14. Biology Project Report on malaria

15. BIOLOGY PROJECT REPORT ON IMMUNE SYSTEM

3

Physical

Education

Work revise all course done till now

https://studyres.com/doc/1861347/investigatory-project-for-class-xii-biology

https://studyres.com/doc/1861347/investigatory-project-for-class-xii-biology

4. Information

Practice

Do all practical question related to sqrl. in practical

5. Physics. Revise all prev. work. Also solve the question sheet given in online class.

6. Chemistry Revise all prev. work and.solve NCERT prob.releted to topic which has

finished.

7. Maths Refer to separate file given by sir

TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777

BASIC CONCEPTS OF CHEMISTRY 1

EXERCISE # 1

Question

based on Laws of chemical combination

::

Q.1 There are two common oxides of Sulphur, one

of which contains 50% O2 by weight, the other

almost exactly 60%. The weights of sulphur

which combine with 1 g of O2 (fixed) are in the

ratio of -

(A) 1 : 1 (B) 2 : 1

(C) 2 : 3 (D) 3 : 2

Sol.[D] First oxide Second oxide

Sulphur 50% 40%

Oxygen 50% 60%

In first oxide 1 parts oxygen combines with

sulphur = 50

50 = 1

Similarly for second oxide = 60

40 = 0.67

So the ratio is 1:0.67 or 3 : 2

Q.2 Iron forms two oxides, in first oxide 56 gram.

Iron is found to be combined with 16 gram

oxygen and in second oxide 112 gram iron is

found to be combined with 48 gram oxygen.

This data satisfy the law of -

(A) Conservation of mass

(B) Reciprocal proportion

(C) Multiple proportion

(D) Combining volume

Sol.[C] For fix mass of Fe i.e. 56 parts, the masses of

oxygen in these two oxides are 16 and 48

respectively, simple ratio is 1 : 3 so it is law of

multiple proportion

Q.3 When 10 ml of propane (gas) is combusted

completely, volume of CO2(g) obtained in

similar condition is -

(A) 10 ml (B) 20 ml

(C) 30 ml (D) 40 ml

Sol.[C] C3H8 + 5 O2 3 CO2 + 4 H2O

Value of CO2 = 3 × 10 ml

= 30 ml

Atomic mass, molecular mass,

formula mass

Question

based on

Q.4 The chloride of a metal contains 71% chlorine

by weight and the vapour density of it is 50.

The atomic mass of the metal will be (valency

of metal is 2) -

(A) 29 (B) 58

(C) 35.5 (D) 71

Sol.[A] Mol. wt. of metal chloride = 2 × V.D.

= 2 × 50 = 100

Let at. wt. of metal is x

so x + 71 = 100

x = 100 – 71 = 29

Q.5 A reaction required three atoms of Mg for two

atoms of N. How many gm of N are required

for 3.6 gm of Mg ?

(A) 2.43 (B) 4.86

(C) 1.4 (D) 4.25

Sol.[C] gm of N = 243

142

× 3.6 = 1.4

Q.6 A hydrated salt of Na2SO3 loses 22.22 % of its

mass on strong heating. The hydrated salt is -

(A) Na2SO3.4H2O (B) Na2SO3.6H2O

(C) Na2SO3.H2O (D) Na2SO3.2H2O

Sol.[D] grams of H2O = 8.77

2.22 × 126

= 35.95

no. of molecules of H2O = 18

95.35 ~ 2

Hence, hydrate is Na2SO3 . 2H2O

Q.7 The average molecular mass of a mixture of

gas containing nitrogen and carbon dioxide is

36. The mixture contain 280 gm of nitrogen,

therefore, the amount of CO2 present in the

mixture is -

(A) 440 gm (B) 44 gm

(C) 0.1mole (D) 880 gm



Sol.[A] Let amount of CO2 persent = x g

44

x

28

280

4444

x28

28

280

36

36 = 44/x10

x280

=

x440

44)x280(

or 15840 + 36 x = 12320 + 44x

or 8x = 3520

x = 8

3520 = 440 g

Q.8 In an ionic compound moles ratio of cation to

anion is 1 : 2. If atomic masses of metal and

non-metal respectively are 138 and 19, then

correct statement is -

(A) molecular mass of compound is 176

(B) formula mass of compound is 176

(C) formula mass of compound is 157

(D) molecular mass of compound is 157

Sol.[B] Metal Cation A (138)

Non-metal Anion B (19)

Ionic compound = AB2

Formula mass (G.M.M. for ionic

compound) = 138 + 2 × 19 = 176

Question

based on Mole-concept

Q.9 The mass of 2 gram atoms of calcium

(Relative atomic mass = 40)

(A) 2 g (B) 0.05 g

(C) 0.5 g (D) 80 g

Sol.[D] mass = 2 × 40 = 80 g

Q.10 Which of the following contains largest number

of atoms -

(A) 1.0 g of O atoms

(B) 1.0 g of O2

(C) 1.0 g of O3

(D) All have equal atoms

Sol.[A] 1.0 g O atoms = NA = 16 g = 1 mole

(B) 32

1 moles

(C) 48

1 moles

(C) 96

1 moles

Q.11 The number of molecules present in 88 g of

CO2 (Relative molecular mass of CO2 = 44)

(A) 1.24 × 1023

(B) 3.01 × 1023

(C) 6.023 × 1024

(D) 1.2046 ×1024

Sol.[D] no. of molecules

= 44

88 × 6.023 × 10

23

= 12.046 × 1023

= 1.2046 × 1024

Q.12 The number of Ca2+

and Cl¯ ions present in

anhydrous CaCl2 is 3.01 × 1023

and 6.023 ×

1023

respectively. The weight of the anhydrous

sample is -

(A) 40 g (B) 55.5 g

(C) 222 g (D) 75.5 g

Sol.[B] Weight of antiydrous sample

= 23

23

1002.6

100.3

× 40 +

23

23

10023.62

10023.6

× 71

= 20 + 35.5 = 55.5 g

Q.13 The largest number of molecules is present in

(A) 34 g of H2O (B) 28 g of CO2

(C) 46 g of CH3OH (D) 54 g of N2O5

Sol.[A] (A) 18

34 × NA = 1.89 NA

(B) 44

28 × NA = 0.64 NA

(C) 32

46 × NA = 1.44 NA

(D) 108

54 × NA = 0.5 NA

Q.14 If NA is Avogadro number, then the number of

valence electrons in 4.2 g of N3–

ions is -

(A) 2.4 NA (B) 4.2 NA

(C) 1.6 NA (D) 3.2 NA

Sol.[A] No. of valence electron = 8 × 14

2.4 × NA

= 2.4 NA

Q.15 Two oxides of a metal contain 50% and 40%

metal M respectively. If the formula of the first

oxide is MO2, the formula of the second oxide

will be -



(A) MO2 (B) MO3

(C) M2O (D) M2O5

Sol.[B] M O2 50 gm ….. 1metal Mx Oy 50% 50%

1gm ……. 50

1 40 60

50 gm ….2oxygen and 40 gm ….. 5

4 metal

1gm …. 50

2 oxygen

For second oxide Atoms of metal

(M) = 50

401 = 0.8 similarly

Atoms of oxygen = 50

602 =

5

12 = 2.4

Hence, ratio of M:O = 0.8 : 2.4 or 1 : 3

Formula of the same metial oxide is MO3

Q.16 Which of the following has the least mass ?

(A) 2 g atoms of nitrogen

(B) 3 ×1023

atoms of carbon

(C) 1 mol of sulphur

(D) 7.0 g of Ag

Sol.[B] (A) 2 × 14 = 28 g

(B) 23

23

10023.6

10312

6g

(C) 1 mole sulphur = 32 g

(D) 7.0 g of Ag = 7 g

Q.17 Insulin contains 3.4% sulphur. What will be

the minimum molecular weight of insulin ?

(A) 94.176 (B) 1884

(C) 941.176 (D) 976

Sol.[C] Minimum mol. wt. of

Insulin = 4.3

100 × 32 = 941.176 g/mole

Q.18 The volume occupied by 7.23 × 1023

molecules

of carbon dioxide and 3.01 × 1023

molecules of

Argon at 0°C and 1 atm pressure is -

(A) 38 mL (B) 3.80 L

(C) 3.8 × 104 mL (D) 3.8 × 10

3 mL

Sol.[C] Volume of CO2 = 23

23

10023.6

1023.7

× 22.4 = 26.8 L

Volume of Argon = 23

23

1002.6

1001.3

× 22.4 = 11.22

Total volume = 26.8 + 11.2 = 38L = 3.8 × 104 mL

Empirical & Molecular formula and

relation in P, V, T and number of moles

Question

based on

Q.19 A carbon compound containing carbon and

oxygen has molar mass equal to 288. On

analysis it is found to contain 50% by mass of

each element. Therefore molecular formula of

the compound is -

(A) C12O9 (B) C4O3

(C) C3O4 (D) C9O12

Sol.[A] %age Moles of elements Molar

element ratio

C 50 12

50 = 4.17

13.3

17.4 = 1.33

O 50 16

50 = 3.13

13.3

13.3 = 1

Hence whole no. of molar ratio of C & O = 4 : 3

So empirical formula = C4O3

empirical formula wt. = 96

n = 96

288 = 3

So mol. formula = 3 × C4O3

= C12O9

Q.20 A sample of impure cuprite, Cu2O, contains

66.6% copper. What is the percentage of pure

Cu2O in the sample -

(A) 75% (B) 25% (C) 60% (D) 80%

Sol.[A] 1 mole copper present in = 6.66

100 × 127 =

190.7g = impure sample

% age purity of Cu2O = 7.190

143 × 100 = 75%

Q.21 A given sample of pure compound contains

9.81 gm of Zn, 1.8 × 1023 atoms of chromium,

and 0.60 mol of oxygen atoms. what is the

simplest formula -

(A) ZnCr2O7 (B) ZnCr2O4

(C) ZnCrO4 (D) ZnCrO6

Sol.[B] moles of Zn = 39.65

81.9 = 0.15

moles of Cr = 23

23

10023.6

108.1

= 0.3

moles of O = 0.6



so whole molar ratio

= 0.15 : 0.3 : 0.6 = 1 : 2 : 4

Hence, formula = ZnCr2O4

Q.22 510 mg of liquid on vaporization in Victor

Meyer's apparatus displaces 67.2 cm3 of air at

(STP). The molecular weight of the liquid is -

(A) 130 (B) 17 (C) 170 (D) 1700

Sol.[C] Mol. wt of liquid

= 2.67

10510 3– × 22400

= 170

Q.23 When 3.2 g S is vapourized at 450ºC and

723 mm pressure, the vapours occupy a volume

of 780 ml. What is the molecular formula of

S vapours under these conditions -

(A) S2 (B) S4 (C) S6 (D) S8

Sol.[D] W = 3.2 g

T = 450 + 273 = 723 K

P = 760

723 = 0.95 atm

R = 0.082 L – atm K–1

mol–1

V = 780 ml = 0.78 L

Now M = PV

WRT =

95.078.0

723082.02.3

= 256 g/mole

So Let formula is Sx

x × 32 = 256

x = 32

256 = 8

Hence, formula = S8

Gravimetric Analysis & limiting Reactant

Question

based on

Q.24 2.76 g of silver carbonate on being strongly

heated yields a residue weighing -

(Ag2CO3

2Ag + CO2 + 2

1O2)

(A) 2.16 g (B) 2.48 g

(C) 2.32 g (D) 2.64 g

Sol.[A] Ag2CO3 Heat

2Ag + CO22

1O2

wt. of residence

= 276

1082 × 2.76 g

= 276

76.2216 g

= 2.16 g

Q.25 1.84 gram mixture of CaCO3 and MgCO3 on

heating gives CO2. Volume of CO2 obtained is

measured to be 448 mL at STP. mass of CaCO3

in mixture is -

(A) 0.5 gram (B) 0.84 gram

(C) 0.92 gram (D) 1.00 gram

Sol.[D] 1.00 gram

Q.26 3 gm of Mg is burnt in a closed vessel

containing 3 gm of oxygen. The weight of

excess reactant left is -

(A) 0.5 gm of oxygen (B) 1.0 gm of oxygen

(C) 1.0 gm of Mg (D) 0.5 gm of Mg

Sol.[B] Mg + 2

1 O2

MgO

3g 3g

= 0.125 moles = 0.1875 moles

Here limiting reactant is Mg

So, oxygen left unreacted is = 0.1875 – 0.125

= 0.0625 moles

Hence, wt of oxygen = 0.0625 × 16 = 1.0 g

Q.27 0.54 gm of metal “M” yields 1.02 gm of its

oxide M2O3. The at. wt. of metal “M” is -

(A) 9 (B) 18

(C) 27 (D) 54

Sol.[C] oxygen

oxygen

metal

metal

E

w

E

w

metalwt.E

54.0 =

8

54.002.1 =

8

48.0

Emetal = 48.0

854.0 = 9

At. wt. of metal

= Valency × Eq. Wt

= 3 × 9

= 27

Q.28 Phosphine (PH3) decomposes to produce P4 (g)

and H2 (g). What would be the change in

volume when 100 ml of PH3 (g) is completely

decomposed ?

(A) 50 ml (B) 500 ml

(C) 75 ml (D) 250 ml



Sol.[C] PH3 4

1 P4(g) +

2

3 H2 (g)

100 mL

volume of P4(g) = 4

100 = 25 mL

volume of H2(g) = 2

3 × 100 = 150 ml

Total volume = 25 + 150 = 175 ml

change in volume = 175 – 100 = 75 ml

Q.29 The minimum quantity in gram of H2S needed

to precipitate 63.5 g of Cu2+

will be -

(Cu2+

+ H2S CuS + 2H+)

Black

(A) 63.5 g (B) 31.75 g

(C) 34 g (D) 2 g

Sol.[C] Cu2+

+ H2S Cu2S + H2

6.35 g

= 1 mole

So 1 mole H2S is required i.e, 34 g

Question

based on Units of concentration

Q.30 25 g of NaOH is dissolved in 50 mL of water.

The molarity of the solution is

(A) 12.5 M (B) 12 M

(C) 24 M (D) 25 M

Sol.[A] Molarity = 40

25 ×

50

1000 = 12.5 M

Q.31 The density of 2.45 M methanol solution in

water is 0.9766 g mL–1

. The molality of the

solution is

(A) 27.3 (B) 2.73

(C) 2.45 (D) 0.273

Sol.[B] wt. of one liter solution

= 1000 × 0.9766 = 976.6 g

wt. of methanol = 2.45 × 32 = 78.4 g

wt. of water = 976.6 – 78.4 = 898.2 g

So molality = 32

4.78 ×

2.898

1000 = 2.73

Q.32 Density of a solution of NaCl in a polar solvent

is 1.2 gram/cc. If % by weight (W/W) of NaCl

in solution is 5.85, then formality and % by

volume (W/V) of NaCl are, respectively, -

(A) 1 and 5.85 (B) 0.1 and 0.585

(C) 1.2 and 7.02 (D) 1.1 and 7.2

Sol.[C] 1.2 and 7.02

Q.33 Density of 3M Na2CO3 solution in water is

1.2 g mL–1

. The percentage by weight and ppm

of Na2CO3 are respectively -

(A) 26.5 and 2.65 × 105

(B) 2.65 and 2.65 × 104

(C) 265 and 2.65 × 103

(D) 5.23 and 2.65 × 106

Sol.[A] Mass of one litre solution = 1.2 × 1000 = 1200 g

wt. of Na2CO3 = 3 × 106 = 318 g

% weight = 1200

318 × 100 = 26.5

and ppm of Na2CO3 = 1200

318× 10

6 = 2.65 × 10

5

Q.34 Mole fraction of methanol in its aqueous

solution is 0.5. The concentration of solution in

terms of percent by mass of methanol is -

(A) 36 (B) 50

(C) 64 (D) 72

Sol.[C] wt. of methanol = 0.5 × 32 = 16 g

wt of water = 0.5 × 18 = 9g

%age mass of motional

= 916

16

× 100

= 25

16 × 100 = 64

Question

based on Eudiometry

::

Q.35 100 ml of CH4 and C2H2 were exploded with

excess of O2. After explosion and cooling, the

mixture was treated with KOH, where a

reduction of 165 ml was observed. Therefore

the composition of the mixture is -

(A) CH4 = 35 ml ; C2H2 = 65 ml

(B) CH4 = 65 ml ; C2H2 = 35 ml

(C) CH4 = 75 ml ; C2H2 = 25 ml

(D) CH4 = 25 ml ; C2H2 = 75 ml

Sol.[A] CH4 + 2O2 CO2 + 2H2O

C2H2 + 2

5 O2 2CO2 + H2O

Let a ml CH4 & b mL C2H2

present in mixture



so a + b = 100 mL …….(1)

and a mL & 2b ml. CO2 are formed

so a + 2b = 165 ml ……… (2)

from eq. (1) & (2)

we get b = 65 mL

and a = 35 mL

Hence, CH4 = 35 mL

& C2H2 = 65 mL

Q.36 A mixture of CO and CO2 having a volume of 20

ml is mixed with x ml of oxygen and electrically

sparked. The volume after explosion is (16 + x)

ml under the same conditions. What would be the

residual volume if 30 ml of the original mixture is

treated with aqueous NaOH ?

(A) 12 ml (B) 10 ml

(C) 9 ml (D) 8 ml

Sol.[A] CO + 2

1 O2 CO2

CO2 + O2 No reaction

Let a ml co & b ml CO2 are present in the

mixture so

a + b = 20 ) × 2 …… (1)

After the explosion a mL CO2

is formed so,

a + b + 2

x = 16 + x

or 2a + 2b – x = 32 ……… (2)

from eq. (1) & (2)

x = 8 ml

Therefore volume of CO in mixture = 8 mL

volume of CO2 = 20 – 8 = 12 ml

+ 2NaOH Na2CO3

CO + NaOH No,

if 30 ml original mixture use then

volume of CO2 in the mixture

= 20

12 × 30 = 18 ml

and volume of CO left unreacted

= 30 – 18 = 12 mL

True or false type questions

Q.37 The mass of carbon present in 36.8g of

potassium ferrocyanide (Mol. mass = 368) is

12g.

Sol. Moles of K4[Fe(CN)]4 = 368

8.36= 0.1 mole

1 mole of K4[Fe(CN)4] give = 40 mole C = 0.4

0.1 ____________= 0.4

n = M

w

n × M = w

0.4 × 12 = 4.4 g

Q.38 A 20% solution of KOH (density = 1.02g/ml)

has molarity = 3.64.

Sol. Am

10dxM

= 64.3

56

1002.120

Fill in the blanks type questions

Q.39 The iron atoms in 1720 amu of ferric

ferrocyanide is .....................

Sol. Fe4 [Fe (CN)6]3

m.w = 7 × 56 + 18 × 26 = 860

no. of molecules = 860

1720 = 2

Fe= 7× 2 = 14

Q.40 The volume of 1.204 × 1024 molecules of water

at 4ºC is ……………

Sol. mole = 2

wt = 36 gm

vol = 36 ml



EXERCISE # 2

Only single correct answer type

questions Part-A

Q.1 Four one litre flasks are separately filled with

the gases hydrogen, helium, oxygen and ozone

at the same room temperature and pressure.

The ratio of total number of atoms of these

gases present in the different flasks would be -

(A) 1 : 1 : 1 : 1 (B) 1 : 2 : 2 : 3

(C) 2 : 1 : 2 : 3 (D) 3 : 2 : 2 : 1

Sol. [C] All gases at NTP have 1 mole = 22.4 litre

22..4 litre have = 1 mole

1 litre have = 4.22

1 mole

2

4.22

NH

A

2

4.22

NHe

A 2

4.22

NO

A

2

34.22

N

O

A

3

So 2 : 1 : 2 : 3

Q.2 The atomic masses of two elements A and B

are 20 and 40 respectively, if x gm of A

contains y atoms, how many atoms are present

in 2x gm of B ?

(A) y (B) 2y

(C) 2

y (D)

4

y

Sol. [A] 20

x× NA = y atom

40

x2× NA =

20

x× NA = y

B also have y atoms

Q.3 If mole percentage of C–12 and C–14 in nature

is 98% and 2% respectively, then the number of

C–14 atoms in 12 g of carbon is -

(A) 1.2 × 1022 (B) 3.01 × 1022

(C) 5.88 × 1023 (D) 6.02 × 1023

Sol.[A] 100g have 2 g

12 g have100

2× 12 ×

14

1× NA ×1

= 1.2 × 1022

atoms

Q.4 A vessel contains 8 gram TiO2, 2.4 gram

carbon and 28.4 gram Cl2. Maximum mass of

TiCl4 which can be produced is -

3TiO2(s) + 4C(s) + 6Cl2(g)

3TiCl4(g) + 2CO2(g) + 2CO(g)

(Consider reaction goes to completion ; atomic

mass of Ti = 48)

(A) 1.90 gram (B) 28.5 gram

(C) 19 gram (D) 38 gram

Sol.[C] TiO2(s) + 4 (s) + 6Cl2(g) 3TiCl4(g) +

2CO2(g) + 2CO(g)

moles taken :- 80

8 = 0.1

12

24 = 0.2

72

4.28 = 0.4

L.R = TiO2 4TiCln produced = 0.1

w = 0.1 × 4TiClM = 0.1 × 190 = 19g

Q.5 Total number of atoms of all elements present

in 1 mole of ammonium dichromate is ?

(A) 14 (B) 19

(C) 6 x 1023 (D) 114 x 1023

Sol.[D] (NH4)2Cr2O7 =19×6.02×1023

= 114 × 1023

atoms

Q.6 25 grams of oleum contains 30% free SO3.

Strength of oleum is -

(A) 130% (B) 106.75 %

(C) 115% (D) 110

Sol.[B] 106.75 %

Q.7 25 gram of A sample of oleum is labelled as

110%. The amount of H2O which should be

added to this sample to get 50 % H2SO4(w/w) is -

(assuming densily of H2O = 1 g/ml)

(A) 25 gram (B) 20 gram

(C) 30 gram (D) 27.5 gram

Sol.[C] 30 gram

Q.8 The number of atoms contained in 11.2 L of

SO2 at N.T.P. are -

(A) 3/2 x 6.02 x 1023 (B) 2 x 6.02 x 1023

(C) 6.02 x 1023 (D) 4 x 6.02 x 1023

Sol.[A] = S + 2O = 3

= 2

3 × 6.02 × 10

23

22.4 litre gas has = 1 mole

1 litre gas has = 4.22

1×11.2 =

2

1mole



Q.9 The vapour density of gas A is four times that

of B. If molecular mass of B is M, then

molecular mass of A is -

(A) M (B) 4M

(C) 4

M (D) 2M

Sol.[B] 4 (V.D.)A = (V.D.)B

(V.D.)B = 2

MB …(1)

(V.D.)A = 2

MA …(2)

4 × 2

MA = 2

MB

MA = 2

MB ×4

2

MA = 4

MB

Q.10 On analysis, a certain compound was found to

contain iodine and oxygen in the ratio of 254

gm of iodine (at. mass 127) and 80 gm oxygen

(at. mass 16). What is the formula of the

compound -

(A) IO (B) I2O

(C) I5O3 (D) I2O5

Sol.[D] Simple ratio of atom = 127

254= 2 for I

Simple ratio of atom = 16

80 = 5 for O

So I2O5

Q.11 The hydrated salt Na2SO4.nH2O, undergoes

55% loss in weight on heating and becomes

anhydrous. The value of n will be

(A) 5 (B) 3 (C) 7 (D) 10

Sol.[D] H2O% = 100

55 =

n18142

n18

= n =10

Q.12 The mass of oxygen that would be required to

produce enough CO, which completely reduces

1.6 kg Fe2O3 (at. mass Fe = 56) is -

(Fe2O3 + 3CO 2Fe + 3CO2)

(A) 240 gm (B) 480 gm

(C) 720 gm (D) 960 gm

Sol. [B] Fe2O3 + O2 Fe2+

+ CO

1

1

E

W=

2

2

E

W

6

160

106.1 3 =

4

32

x

x = 480 gm

Q.13 12g of Mg (atm. mass 24) will react completely

with acid to give -

(A) One mol of H2

(B) 1/2 mol of H2

(C) 2/3 mol of O2

(D) Both 1/2 mol of H2 and 1/2 mol of O2

Sol.[B] Mg + 2HCl MgCl2 + H2

1 2 1 1

2

1 1

2

1

2

1 H2

Q.14 If one mole of ethanol (C2H5OH) completely

burns to carbon dioxide and water, the weight

of carbon dioxide formed is about -

(C2H5OH + 3O2 2CO2 + 3H2O)

(A) 22g (B) 45g (C) 66g (D) 88g

Sol. [D] C2H5OH + 3O2 2CO2 + 3H2O

1 3 2×44 3

= 88 g

Q.15 Calculate the weight of lime (CaO) obtained by

heating 200 kg of 95% pure lime stone

(CaCO3).

(A) 104.4 kg (B) 105.4 kg

(C) 212.8 kg (D) 106.4 kg

Sol.[D] CaCO3 CaO + CO2

100 g 56 g 44 g

100 g CaCO3 give = 56g CaO

1 g CaCO3 give = 100

56

200×100

95gCaCO3 give =

100

56×

95200

100

=106.4 kg

Q.16 The mass of 70% H2SO4 required for

neutralisation of 1 mol of NaOH

(A) 49 gm (B) 98 gm

(C) 70 gm (D) 34.3 gm



Sol.[C] 2

2

1

1

E

w

E

w =

49

100

70x

= 1 × 1

x = 70

10049 = 70 gm

Q.17 12 litre of H2 and 11.2 litre of Cl2 are mixed

and exploded. The composition by volume of

mixture is -

(A) 24 litre of HCl

(B) 0.8 litre Cl2 and 20.8 lit HCl.

(C) 0.8 litre H2 & 22.4 litre HCl

(D) 22.4 litre HCl

Sol. [C] H2 + Cl2 2HCl

1 mole 1 mole 2 mole

11.2 litre at NTP gas has = 0.5 mole Cl2

11.2 = 12 litre NTP = 0.5 mole H2

0.8 litre H2 = remaining

1 mole HCl = 22.4 litre

Q.18 What volume of 0.4-M FeCl3.6H2O will

contain 600 mg of Fe3+ ?

(A) 49.85 mL (B) 26.78 mL

(C) 147.55 mL (D) 87.65 mL

Sol.[B] Molarity = )ml(V

1000n

V=M

1000n=

564.0

10100010600 3

= 26.78 ml

Q.19 The mole fraction of a given sample of I2 in

C6H6 is 0.2. The molality of I2 in C6H6 is -

(A) 0.32 (B) 3.2

(C) 0.032 (D) 0.48

Sol.[B] Mole fraction of I2 = XA

XA =21

1

nn

n

XB =

21

2

nn

n

XA + XB = 1

XA = 0.2

XB = 0.8

B

A

X

X=

2

1

n

n

n1 : n2 = 1 : 4

)g(784MnwM

wn

solventofMass

2

2

Molality = )kgin(solventofmass

n1

= 784

1

× 1000 = 3.21 m

Q.20 1 mol of N2 and 4 mol of H2 are allowed to

react in a vessel and after reaction, H2O is

added. Aqueous solution required 1 mol of

HCl. Mol fraction of H2 in the remaining

gaseous mixture after reaction is -

(A) 6

1 (B)

6

5

(C) 3

1 (D) None of these

Sol. [B] N2 + 3H2 2NH3

1 3 2

2

1

2

3 1

1–2

1 4 –

2

3

2

1

2

5

1mole HCl required 1 mol NH3 for

neutralization

Mole fraction of H2 = 21

1

nn

n

=

2

5

2

12

5

= 6

5

Q.21 Calculate the weight of BaCl2 needed to

prepare 250 mL of a solution having the same

concentration of Cl– ions as in a solution of

KCl of concentration 80 g/L. (Ba = 137.4, Cl =

35.5) -

(A) 27.92 g (B) 14.50 g

(C) 22.52 g (D) 11.46 g

Sol.[A] Concentration of Cl– of BaCl2 = Cl

– of KCl

2Cl– = Cl

–

S = 80 g/L

BaCl2 = 74.5



Moles = M

S =

5.74

80

1 litre solution contain = 5.74

80

4

1 litre solution contain =

5.74

80×

4

1

45.74

80

=

207

x

93.27x

Q.22 0.2 mole of HCl and 0.1 mole of barium

chloride were dissolved in water to produce a

500-mL solution. The molarity of the Cl– ions is -

(A) 0.06 M (B) 0.09 M

(C) 0.12 M (D) 0.80 M

Sol.[D] HCl Cl

0.2 mole

BaCl2 2 Cl–

2 × 0.1 = 0.2

Total moles of Cl– = 0.4

M = vm

1000w

Molarity = 500

10004.0 = 0.8 4.0

m

w

Q.23 A sample of H2SO4 (density 1.8 g/ml) is 90%

by weight. What is the volume of the acid that

has to be used to make 1 litre of 0.2-M H2SO4?

(A) 16 mL (B) 10 mL

(C) 12 mL (D) 18 mL

Sol.[C] M = AM

10dx =

98

108.190 = 16.53

16.53 M solution contain = 1 litre = 1000 ml

1 M solution contain = 53.16

1000

0.2 M solution contain =53.16

1000× 0.2 = 12 ml

Q.24 A sample of Na2CO3. H2O weighing 1.24 g is

added to 200 mL of a 0.1-N H2SO4 solution.

The resulting solution becomes -

(A) acidic (B) strongly acidic

(C) alkaline (D) neutral

Sol.[D] No. of equivalent of Na2CO3.H2O = No. of

equi. of H2SO4

N = E

w×

V

1000 = 0.1 N

=

2

124

24.1×

200

1000 = 0.1 N

So neutral solution

Q.25 125 mL of 10% NaOH (w/V) is added to 125

mL of 10% HCl (w/V). The resultant solution

becomes-

(A) alkaline (B) strongly alkaline

(C) acidic (D) neutral

Sol.[C] No. of equivalents of acid (HCl) is greater than

No. of equ. Base so that solution is acidic.

No. of equivalents of Base = N1V1

V1 = 125 ml

N1 = E

w×

100

1000

= 40

10×

100

1000 = 2.5

N1V1 = 2.5 × 125

No. of equivalents of acid = N2V2

V2 = 125

N2 =5.35

10×

100

1000

= 5.35

100= 2.82

N2V2 = 2.82 × 125

1122 VNVN

Q.26 Equal volumes of 0.50 M of HCl, 0.25 M of

NaOH and 0.75 M of NaCl are mixed. The

molarity of the NaCl solution is -

(A) 0.75 M (B) 1/3 M

(C) 0.50 M (D) 2.00 M

Sol.[B] HCl + NaOH NaCl

0.5 0.25 0.25

N1V1 = 0.25 × V

N2V2 = 0.75 × V

= )VV(

VNVN

21

2211

= VV2

V75.0V25.0

=

3

1N



N = n × M

3

1× 1 = M

M = 3

1M

Q.27 How many grams of copper will be replaced in

2 L of a 1.50-M CuSO4 solution if the latter is

made to react with 27.0 g of aluminium ?

(Cu = 63.5, Al =27.0)

(3CuSO4 + 2Al Al2(SO4)3 + 3Cu)

(A) 190 g (B) 95.25 g

(C) 48 g (D) 10 g

Sol.[B] 2

1

W

W=

2

1

E

E

27

W1 =27

3

2

5.63

W1 = 63.5 × 2

3×

27

27= 95.25 g

Q.28 The ratio of the molar amounts of H2S needed

to precipitate the metal ions from 20 ml each

of 1M Cd(NO3)2 and 0.5 M CuSO4 is-

(Cd(NO3)2 + H2S CdS + 2HNO3)

(A) 1 : 1 (B) 2 : 1

(C) 1 : 2 (D) Indefinite

Sol.[B] Cd2+

+ Cu2+

1× 2 0.5 × 2

2 mole 1 mole

2 : 1

Q.29 In which mode of expression, the concentration

of a solution remains independent of

temperature -

(A) Molarity (B) % w/V

(C) Formality (D) Molality

Sol.[D] m = solvent

solute

w

n & n and w are independent of

temperature.

Q.30 The largest number of molecules is in -

(B.H.U. 1997)

(A) 36 g of water (B) 28 g of CO2

(C) 46 g of CH3OH (D) 58 g of N2O5

Sol.[A] 36gH2O = 18

36= 2 mole H2O

28g CO2 = 44

28 =

11

7mole

46g CH3OH = 32

46 =

16

23

58g N2O5 = 108

58 =

54

29

Q.31 Which has the highest weight?(A.F.M.C. 1997)

(A) 1 m3 water

(B) 10 litre of Hg (density of Hg = 13.6 gm/mL)

(C) A normal adult man

(D) All have same weight

Sol.[A] 1m3 H2O = 1000 kg

10HgO = 136 kg

About = 70 – 80 kg

Q.32 The number of atoms of Cr and O are 4.8 × 1010

and 9.6 × 1010 respectively. Its empirical

formula is - (CPMT 1997)

(A) Cr2O3 (B) CrO2

(C) Cr2O4 (D) None of these

Sol.[B] E.F. = 10108.4Cr

10106.3O

= CrO2.

Q.33 Volume of a gas at NTP is 1.12 × 10–7 c.c.

The number of molecules in it will be -

(B.H.U. 1997)

(A) 3.1 × 1020 (B) 3.01 × 1012

(C) 30.1 × 1023 (D) 3.01 × 1024

Sol.[B] 6.023 × 1023

molecules 22.4 = 22.4 × 103 c.c

1 cc = 3

23

104.22

10023.6

molecule

1.12 × 10–7

cc = 1.12 × 10–7

×

3

23

104.22

10023.6

= 3.01 × 10

12

Q.34 Haemoglobin contains 0.33% of iron by

weight. The molecular weight of haemoglobin

is approximately 67200. The number of iron

atoms (at wt. of Fe = 56) present in one

molecule of haemoglobin are -

(A) 1 (B) 2

(C) 4 (D) 6

Sol.[C] Let Fe be there



67200

56n × 100 = 0.33 n =3.96 ~ 4

One or more than one correct

answer type questions Part-B

Q.35 1 mol BaF2 + 2 mol H2SO4 resulting

mixture will be neutralised by ?

(A) 1 mol of KOH (B) 2 mol of Ca(OH)2

(C) 4 mol KOH (D) 2 mol of KOH

Sol.[B,C]BaF2 + H2SO4 BaSO4 + 2HF

No. of equi. No. of equi. 0

0

1 × 2 2 × 2

2 4 2 2

0 2 2 2

H2SO4 + Ca(OH)2 Complete neutralization

2 × 2 2 × 2

4 4

H2SO4 + KOH Complete neutralization

4 mol

2 × 2 4 × 1

4 4

Q.36 11.2 g of mixture of MCl (volatile) and NaCl

gave 28.7 g of white ppt with excess of AgNO3

solution. 11.2 g of same mixture on heating

gave a gas that on passing into AgNO3

solution gave 14.35 g of white ppt. Hence ?

(A) Ionic mass of M+ is 18

(B) Mixture has equal mol fraction of MCl and

NaCl

(C) MCl and NaCl are in 1 : 2 molar ratio (D) Ionic mass of M+ is 10

Sol.[A,B]MCl + NaCl + Ag NO3 15y ………(1)

x (11.2 – x) 28.7

MC Cl2 + Ag NO3 AgCl.

14.35 gm while ppt.

5.35y

x

=

5.143

35.141 = 1 From

or x = 1 (y + 35.5)

Substituting the value of x in Eq = 2

)5.35y(

)5.35y(1

+

5.58

x–2.11 = 2

1 + 5.58

x–2.11 =2 or x = 5 .35

y = 18

From above data mixture has equal mole

fraction of MCl and NaCl

Q.37 Which of the following has same percentage of

carbon as in ethane -

(A) 2-Butene (B) Cyclohexane

(C) Cyclohexene (D) 2-Methyl but-2-ene

Sol.[A,B,D]

CH3–CH=CH–CH3

Total weight = 56

Carbon% = 56

48× 100 = 86%

Same percentage in cyclohexane & 2-methyl

but-2-ene

Q.38 10 g of a sample of silver which is

contaminated with silver sulphide produced

11.2 mL of hydrogen sulphide at S.T.P. by

treatment with excess of hydrochloric acid. The

mass of silver sulphide in the sample is -

(Ag = 108; S = 32)

(A) 1.24 g (B) 124 mg

(C) 5 × 10–4 mol (D) 62 g

Sol.[B,C] AgS HCl2

H2S + 2AgCl

11.2 mL

at STP

(a) = 22400

1× 11.2 = 0.5 × 10

–3 mole of H2S

(a) Low of equivalence

0.5 × 10–3

= 248

x

x = 124 g

Q.39 2.0 g of a tri-atomic gaseous element was found

to occupy a volume of 448ml at 76cm of Hg

and 273K. The mass of its atom is ?

(A) 33.3 amu (B) 5.53 × 10–23 g

(C) 33.3 g (D) 5.53 amu

Sol.[A,B] n = M

w =

4.22

V

x3

2 =

4.22

448

x = 33.33 atom

33.33 × 1.6 × 10–24

gram



= 5.53 × 10–23

gm

Q.40 2.4 g of pure Mg (at. mass 24) is dropped in

100ml of 1N HCl. What is true ?

(A) 1.122 L of hydrogen is produced at S.T.P.

(B) 0.01 mol of magnesium is behind

(C) 0.1 mol of Mg+ ions are formed in

solution

(D) HCl is the limiting reagent

Sol.[A,D]

Mg + 2HCl Mg Cl2 + H2

2.4/24 (.1 mole) .1 mole

100 ml, 1M m mole .05 mole

5 mole

Consumed 100 × 10

–3

= 1 mole

1 2

HCl is limiting reagent its 0.1 mole will be

consumed. Hence 0.5 mole Mg2+

ions will be

used. i.e. .05 mole 1.12 L Hydrogen will be

produced

Assertion-Reason type questions Part-C

Choose any one of the following four

responses.

(A) If both Assertion and Reason are true

and the Reason is correct explanation of

the Assertion.

(B) If both Assertion and Reason are true

but the Reason is not correct

explanation of the Assertion.

(C) If Assertion is true but the Reason is

false.

(D) If Assertion is false & Reason is true.

Q.41 Assertion : 100 ml of 20% v

w NaOH solution

when mixed with 60 gm of 40% H2SO4,

obtained mixture is alkaline in nature.

Reason : Resulting mixture has pH > 7 (at 25ºC)

Q.42 Assertion : Molality (M) of a solution is

related with molarity (m) of solution as

1000

M

m

1

d

1

M

1 1

Here, d = density of solution in gm/mL

M1 = molar mass of solute.

Reason : Molality of a solution is equal to

number of moles of solute dissolved in 1 kg of

solution.

Q.43 Assertion : Vapour density of sulphur vapour

relative to oxygen is 2 because sulphur atom is

twice as heavy as that of O atom.

Reason : Vapour density depends upon the

molecular state of the substance.

Sol.[D] Molecular formula of sulphur is S8.

Q.44 Assertion : Equal volumes of all the gases

contain equal number of atoms.

Reason : Atom is the smallest particle which

takes part in chemical reactions.

Sol.[D] Equal vol contain equal molecules at same P.T.

Column Matching type questions Part-D Q.45

(A)

Concentration units

independent of temperature

variation

(i) Molality

(B)

Concentration units

dependent on temperature

variation

(ii) Molarity

(C)mg of solute present in 1 kg

of solution represent(iii)

ppm

concentration

(D)moles of solute present in 1

kg of solvent represent(iv)

moles of

solute in

hundred moles

of solution

Column-A Column-B

Sol. A (i, iii, iv) B (ii) C (iii), D (i)

Molality moles of solute in 1000 gm solvent thus

independent of temp.

molarity depends on temp.

Mg of solute in 1kg of solution = ppm concentration



Q.46

(A)

On decomposition 50%

increase in volume at the same

temperature and pressure

(i) NH3

(B)

On decomposition 100%

increase in volume at the same


(ii) HCl

(C)

On decomposition no change

in volume at the same


(iii) O3

(D) Gases highly soluble in water (iv) HBr

Column-A Column-B

Sol. A (iii) B (i, iv) C (ii), D (i, ii, iv)

2 NH3 N2 + 3H2

(at 50% decomposition volume increase )

2 HCl H2 + Cl2

NH3, HCl and COCl2 are polar thus highly

soluble.

2HCl H2 + Cl2

no effect of pressure. thus no

change in volume

EXERCISE # 3

Subjective Type Questions Part-A

Q.1 An alloy of Iron (54.7%), nickel (45.0%) and

manganese (0.3%) has a density of 8.17 g cm–3.

How many iron atoms are there in a block of

alloy measuring 10.0 cm x 20.0 cm x 15.0 cm ?

Sol. = V

w

Volume of block = a3

= 10 × 20 × 15 cm3

= 8.17 g/cm3

w = × V

= 100

15201017.8

= 24510 g

Molecules of Fe = AN

FeofMole

= 10056

7.5424510

= 1.44 × 1026

atoms

Q.2 Equal masses of oxygen, hydrogen and

methane are taken in a container in identical

conditions. Find the ratio of the volume of the

gases.

Sol. At NTP or STP 1mole all gases have 22.4 litre

volume

2On =

32

w :

2Hn = 2

w :

4CHn = 16

w

32

w :

2

w :

16

w

1 : 16 : 2

Q.3 A given mixture consists only of pure

substance X and pure substance Y. The total

weight of the mixture is 3.72 gm. The total

number of moles is 0.06. If the weight of one

mole Y is 48 gm and if there is 0.02 mole X in the

mixture, what is the weight of one mole of X ?

Sol. Total mole = 0.06

Moles of x = 0.02

Moles of y = 0.06 – 0.02 = 0.04

1 mole of y have = 48

0.04 mole of y have = 48 × 0.04 = 1.92 g

Total weight of mixture = 3.72

Weight of x = 3.72 – 1.92 = 1.80

n = M

w

M = n

w=

02.0

80.1= 90 g

Q.4 In a certain region of space there are only

5 molecules per cm3 on an average. The

temperature is 3 K. What is the average

pressure of this very dilute gas ?

Sol. PV = nRT

P = MV

w

RT



HerenV

w

= 5 molecule /cm

3 = 5×10

3 L

or R = 0.0821 L-atom/K-mole = 0.0821×103

P = 23

3

1002.6

30821.0105

= 2.04 × 10

–21 atom

Q.5 A mixture of CuO and Cu2O contain 88%. Cu.

What is the percentage of CuO present in the

mixture ?

Sol. Mixture of CuO & Cu2O contains 88% Cu

i.e. 165.63

x

+

165.632

x–100

= 88

x = 9.06 %

Q.6 A sample of clay was partially dried and then

contained 50% silica and 7% water. The

original clay contained 12% water. Find the %

of silica in original sample.

Sol. 7% water present in = 50% silica

1 % water present in = 7

50

12% water present in = 7

50×12×

100

57

= 47.31%

Q.7 A drug mariguana owes its activity to terahydro

cannabinol, which contains 70% as many as

carbon atoms as hydrogen atoms and 15 times

as many hydrogen atoms as oxygen atoms. The

number of moles in a gram of terahydro

cannabinol is 0.00318. Determine its molecular

formula.

Sol. From question

C 0.7 × 15

H – 15

0 – 1

or C10.5 H15 O1

empirical formula

i.e. C21 H3O O2

Q.8 Zinc metal reacts with hydrochloric acid by the

following reaction:

Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2(g) if

0.30 mol Zn is added to hydrochloric acid

containing 0.52 mole HCl. How many moles of

H2 produced.

Sol. Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

Produced H2 gas depends only HCl so HCl

moles determined the moves of H2.

2 mole HCl produced = 1mole H2

0.52 mole HCl produced = 2

1 × 0.52

= 0.26 mole of H2

Q.9 NaBr, used to produce AgBr for use in

photography can itself be prepared as follows :

Fe + Br2 FeBr2

FeBr2 + Br2 Fe3Br8

(not balanced)

Fe3Br8 + Na2CO3 NaBr + CO2 + Fe3O4

(not balanced)

How much Fe in kg is consumed to produce

2.50 × 103 kg NaBr.

Sol. (given NaBr = 2.50 × 10–3

Kg = 2.50 g).

Fe + Br2 FeBr2

3FeBr2 + Br2 Fe3Br8

Fe3Br8 + 4 Na2CO3 8NaBr + 4 CO2 + Fe3O4

8 mole NaBr obtained from = 1 mole Fe3Br8

1 mole NaBr obtained from = 8

1

103

50.2 mole mole NaBr obtained from =

8

1×

103

50.2

= 0.00s32 Fe3Br8

So 1 mole Fe3Br8 contain = 3 mole Fe

0.0032 mole Fe3Br8 contain = 3 × 0.0032

= 0.0096 mole Fe

So weight = 0.0096 × 56 × 103 = 5047 kg

Q.10 A compound which contains one atom of X and

two atoms of Y for each three atoms of Z is

made by mixing 5.00 g of X, 1.15 x 1023 atoms

of Y and 0.03 mole of Z atoms. Given that only

4.40 g of compound results. Calculate the

atomic weight of Y if the atomic weight of X

and Z are 60 and 80 a.m.u. respectively.

Sol. Emprical formula = XY2Z3

According to the Q.

X + 2Y + 3Z XY2Z3

5.00 gm 1.15×1023

atom 0.03 moles



or 60

5moles

23

23

10023.6

1015.1

mole 0.03 moles

or 0.083 mole 0.191 mole 0.03 moles

It is clear that here Z is limiting reagent

hence X + 2Y + 3Z XY2Z3

0.03 0.01 mole

According to the condition,

0.01 mole = 4.40 gm

1 mole = 440 g

given that atomic wt. of X & Z are 60 & 80

amu respectively

hence

60 + 2y + 3 × 80 = 440

or 2y + 60 + 240 = 440

or 2y = 440 –300

or y = 2

140=70

Atomic wt. of y = 70 amu.

Q.11 7.5 ml of a gaseous hydrocarbon was exploded

with 36 ml of oxygen. The volume of gases on

cooling was found to be 28.5 ml. 15 ml of

which was absorbed by KOH and the rest was

absorbed in solution of alkaline pyrogallol. If

all volumes are measured under same conditions,

deduce the formula of the hydrocarbon.

Sol. CxHy + (2 + y/4) O2 xCO2 + y/2 H2O

1 mole

4

yx mole x mole

1 ml

4

yx mole x ml

Only O2 reacts with KOH so produce O2 is

15 ml.

1 ml CxHy produce = x ml CO2

7.5 ml __________ = 7.5 × x ml CO2

7.5 × x = 15

x = 2

1 ml CxHy required ______ = (x + y/4) ml O2

7.4 _______________ = (x + y/4) × 7.5

remaining O2 = 36 – (x + y/4) × 7.5

= 36 – (x + y/4) × 7.5 + 15 = 28.5

= 36 – 15 – 4

5.7 × y + 15 = 28.5

= 36 – 28.5 = 4

5.7 × y

= 7.5 = 4

5.7 × y y = 4

So hydrocarbon C2H4.

Q.12 10 ml of a mixture of CH4, C2H4 and CO2 were

exploded with excess of air. After explosion,

there was contraction on cooling of 17 ml and

after treatment with KOH, there was further

reduction 14 ml. What is composition of

mixture ?

Sol. CH4 + C2H4 + CO2 = 10 ml

a + b + c = 10 ml ...(1)

After explosion, there was contraction

Volume of mixture + volume of O2 – volume of

CO2 formed = 17

a + b + c + 2a + 3b – [(a + 2b + 10 – (a + b)]

= 17

3a + 3b + c = 27 …(2)

Volume of formed CO2 = observed KOH

a + 2b + 10 – (a + b) = 14

CH4+ 2O2 CO2 + 2H2O

a 2a 0 0

a

C2H4 + 3O2 2CO2 + 2H2O

b 3b 0 0

2b

Solving equation (1), (2) and (3) and we will

get

a = 4.5 ml(CH4), b = 4 ml(C2H4),

c = 1.5 ml(CO2)

Q.13 Water is evaporated from 135 mL of 0.224 M

MgSO4 solution until the solution volume

becomes 105 mL. What is the molarity of

MgSO4 in the solution that results ?

Sol. V1 = 135 mL

M1 = 0.224 M

V2 = 105 mL

M2 = 2

M1V1 = M2V2

M2 = 105

224.0135 = 0.288 M



Q.14 (a) How much Ca(NO3)2 in mg must be

present in 50 mL of a solution with 2.35

ppm of Ca? (d = 1 gm/cc)

(b) What is the molarity of NaCl solution

having 1.52 ppm of Na and the solution

density is 1.0 g/cm3?

Sol. (a) 2.35 ppm means

106 gm of water contains = 2.35 gm of Ca

or 106 ml of water contains = 2.35 gm of Ca

(dwater = 1 g/cc)

1 ml water contains = 610

35.2

50 ml water contains = 2.35 × 10–6

× 50

or 50 ml water contains = 117.5 × 10–6

gm

No. of moles of Ca

in 50 ml of water = 40

105.117 6

= 2.9375 × 10–6

moles

Hence there should be 2.9375 × 10–6

moles

of Ca(NO3)2 in 50 ml water

Wt. of Ca(NO3)2 in 50 ml water

= No. of moles × Mol. Wt.

= 2.9375 × 10–6

× (164)

= 481 × 10–6

gm

= 481 × 10–6

× 103 mg

= 481 × 10–3

mg = 0.48mg

(b) It is clear that

106 gm of water contains 1.52 gm of Na

Given that dw = 1 gm / cc hence

106 ml of water contains 1.52 gm of Na

1 ml of water contains 610

52.1 gm

1000 ml of water contains 1.52 × 10–6

×1000

1 L of water contains _ 1.52 × 10–3

gm of Ca

No. of mole of Na = wt.At

gramin.wt=

23

1052.1 3–

= 0.066087×10–3

moles

hence there are 6.6087×10–5

mole of Na or

NaCl in 1 litre water

The molarity of solution = 6.6087×10–5

M NaCl

Q.15 Calculate the ionic strength (in gm/litre)of a

solution containing 0.2 M NaCl and 0.1 M

Na2SO4.

Sol. NaCl Na+ + Cl

–

0.2 M 0.2 M 0.2 M

Na2SO4 2Na+ + SO4

–2

0.1 M 2 × 0.1 M 0.1 M

= 0.2 M

Ionic strength of Na+

in solution

= 0.2 × 23 + 0.2 × 23

= 9.2 g/L

ionic strength of Cl–

= 0.2 × 35.5

= 7.10 g/L

Ionic strength of SO4–2

= 0.1 × 96

= 9.6 g/L

Q.16 Determine the volume of dilute nitric acid

(d = 1.11 g mL–1, 19% w / w HNO3) that can

be prepared by diluting with water 50 mL of

conc. HNO3 (d = 1.42 g mL–1, 69.8% w / w).

Sol. M1V1 = M2V2

M1 = Am

10dx

M1 = 63

1011.119 , V1 = ?

M2 = 63

1042.18.69 , V2 = 50 ml

V1 = 101011.11963

63501042.18.69

V1 = 234.98 ml.

Q.17 A mixture of FeO and Fe3O4 when heated in air

to a constant weight gains 5% in its weight.

Find the composition of the initial mixture.

(Fe = 56, O = 16)

Sol. FeO + Fe3O4 Initial = 100 g

72

xgm

232

x100 gm

4

1 O2

4

1 O2 After heating

in air

Fe2O3 Fe2O3 After heating

= 105 gm



32243

322

OFe3O2

1OFe2&

OFeO2

1FeO2

g5100105Oxygen

of.wttheisThere

Hence 1 mole of FeO reacts with 4

1 mole of O2

& 1 mole of Fe3O4 reacts with 4

1 mole of O2

According to the question

232

x100

4

1

72

x

4

132 = 5

or 4

1×

72

x× 32 +

4

1×

232

x100× 32 = 5

or 9

x+

29

x100= 5

929

x9900x29

=5

or 20x + 900 = 5 × 29 × 9

20x = 1305 – 900

or 20x = 405 x = 20

405= 20.25

Hence the quantity of FeO in mixture is

20.25%

Hence the quantity of Fe3O4 in mixture is

100 – 20.25 = 79.75 %

Q.18 1 g of a sample containing, NaCl, NaBr and

inert material, with excess of AgNO3, produces

0.526 g of precipitate of AgCl and AgBr. By

heating this precipitate, in a current of chlorine,

AgBr converted to AgCl and the precipitate

then weighted 0.426 g. Find the percentage of

NaCl and NaBr in the sample.

Sol.

NaCl + NaBr

AgNO3

AgCl + AgBr 0.526 gm

AgCl + AgCl 0.426 gm

= 0.100 g (difference)

AgBr Cl

AgCl

188 143.5

= 188 – 143.5 = 44.5

1 mole AgCl convert in AgCl = changing in

mass = 44.5

= 44.5 gram AgBr has = 1 mole

1 gram AgBr has = 5.44

1

0.1 gram AgBr has = 5.44

1 × 0.1

= 0.00224 mole of AgBr

1 mole AgBr require = 1 mole of NaBr

So 0.00224 mole = 0.002241 × 103

= 0.2314 × 100

= 23.1% of NaBr

0.002241 mole AgBr have = 0.002241 × 188

= 0.421 g

So AgCl = 0.526 – 0.42

= 0.103 g of AgCl

1 mole AgCl requires = 1 mole NaCl

= 5.143

103 ______________ =

5.143

103 × 58.5

= 0.419 × 100

= 4.19 % NaCl

Q.19 A drop (0.5 mL) of 12.0 M HCl is spread over

a sheet of thin aluminium foil. Assuming that

all the acid dissolves through the foil, what will

be the area, in cm2, of the hole produced?

(Density of Al = 2.70 g cm-3 ; thickness of the

foil = 0.10 mm)

Sol. Meq. of Al = Meq. of HCl

= 12 × 0.05 = 0.6

Weight of Al = 1000

96.0 = 0.0054 g

meq = M × V

Volume of Al foil = 7.2

0054.0 mL or cm

3

= 0.002 cm3

Now, Area × thickness = volume

Area = 01.0

002.0 = 0.2 cm

2

Q.20 By the reaction of carbon and oxygen, a

mixture of CO and CO2 is obtained. What is

the composition of the mixture obtained when

20 grams of O2 reacts with 12 grams of carbon?



Sol. C + 2

1O2 CO

1 mole 2

1 mole 1 mole

So O2 is limiting reagent

2

1 mole O2 produced = 1 mole CO

1 mole O2 produced =

2

1

1 = 2

32

201 mole O2 produced = 2 ×

32

20 = 1.25 mole

C + O2 CO2

1 mole 1mole 1 mole

1 mole O2 produced = 1 mole CO2

32

20 mole O2 produced = 1 ×

32

20 = 0.63

So composition of mixture

CO = 63.025.1

25.1

× 100 = 65.66 %

CO2 = 63.025.1

63.0

× 100 = 34.34 %

Q.21 One litre of milk weighs 1.035 kg. The butter

fat is 4% in volume of milk has density of 875

kg/m3. Find the density of fat free skimed milk

Sol. 1 litre milk = 1.035 kg

100 litre milk = 1.035 × 100 = 103.5 kg

1 L volume have density = 0.875

4 L volume have density = 3.5

4% fat

1 L milk have density = 1.035

100 milk have density = 100

= 96

100

= 1.041 × 1000 log/m3

= 1041.16 kg/m3

Q.22 One mole of a mixture of CO and CO2 requires

exactly 20 gram of NaOH in solution for

complete conversion of all the CO2 into

Na2CO3. How many grams more of NaOH

would it require for conversion into Na2CO3 if

the mixture (one mole) is completely oxidised

to CO2 ?

Sol. 2NaOH + CO2 Na2CO3 + H2O

2 mole 1 mole 1 mole

40

20 =

2

1

2

1 mole

4

1 mole

Moles of CO = 1 – 4

1 =

4

3

So CO convert in CO2

CO+2

CO2+4

= 4 – 2 = 2

n factor = 2

4

3 × 2 =

40

w

w = 60 g

Q.23 Polyethylene can be produced from CaC2

according to the following sequence of

reactions

CaC2 + H2O CaO + C2H2

C2H2 + H2 C2H4

nC2H4 (CH2CH2)n

Calculate mass of polyethylene which can be

produced from 20 kg of CaC2. % Yield of each

step is 50%.

Sol. CaC2 + H2O CaO + C2H2

C2H2 + H2 C2

nC2H4 (CH2CH2)n

By POAC C2in CaC2 %50 C2 in C2H4

%50 C2 in (CH2CH2)n

)CHCH( 22n = 0.5 × 0.5 × 0.5 ×

64

100020

0.0390625

22CHCHW = 0.039025 (24 + 4) = 1.09375 kg

Q.24 A particular 100-octane aviation gasoline used

1 cc of (C2H5)4Pb, of density 1.66 gm/c.c, per

litre of gasoline. (C2H5)4Pb is made as follows :

4 C2H5Cl + 4NaPb (C2H5)4Pb + 4NaCl

How many gram of C2H5Cl is needed to make

enough (C2H5)4Pb for 10 litre of gasoline.

(atomic mass of Pb = 206)

Sol. 1 litre gasoline require 1cc of (C2H5)4 Pb

10 = 10 cc 16.6 g 322

6.16 mole



moles of C2H5Cl veq = 4 × 322

6.16moles

ClHC 52w = 4 ×

322

6.16 (24 + 3 + 35.5) = 13.3

Q.25 The plastic industry uses large amounts of

phthalic anhydride, C8H4O3, made by the

controlled oxidation of naphthalene. (C10H8)

2 C10H8 + 9O2 2 C8H4O3 + 4 CO2 + 4H2O

Since some of the naphthalene is oxidized to

other products, only 70% of the maximum

yield predicted by the equation is actually

obtained. How much phthalic anhydride would

be produced in practice by the oxidation of 200

gm of C10H8 ?

Sol. 200 gm C10H8 81012

200

=

16

25 moles

348 OHCn produced = 0.7 ×

16

25

348 OHCw = 0.7 ×

16

25 × (8 × 12 + 4 × 1 + 3 ×

16) = 161.875

Passage based objective questions Part-B

Passage-1 (Question 26 to 28) :

Benzamin franklin did an experiment to estimate

molecular size and Avogadros number. He spreaded

one tea spoon of oil on water. Volume of oil

franklin used was 4.9 c.c and the area covered by

oil was 2.0 × 107 cm

2. Density of oil is 0.95 g/c.c

and molar mass of oil is 200 gram. It is assumed

that the oil molecules are tiny cubes that pack

closely together and form a layer only one molecule

thick.

Q.26 Length of the side of one molecule is -

(A) 1.7 cm (B) 1.7 ×10–7

cm

(C) 2.45 ×10–7

cm (D) 1.225 × 10–7

cm

Sol.[C] 2.45 ×10–7

cm

Q.27 Number of molecules present in one teaspoon

of oil is -

(A) 3.33 ×1020

(B) 3.33 × 1023

(C) 6.02 × 1023

(D) 1.67 × 1020

Sol.[A] 3.33 ×1020

Q.28 Value of Avogadro's number calculated by

Benzamin's experiment is -

(A) 6.02 × 1023

(B) 1.43 × 1022

(C) 6.02 × 1022

(D) 1.43 × 1023

Sol.[B] 1.43 × 1022


Questions given below are based on two industrial

observations :

(i) Phosphoric acid H3PO4, is widely used to

make fertiliser & can be prepared by a two

step process

Step I : P4 + 5O2 P4O10

Step II : P4 O10 + 6H2O 4H3PO4

We allow 310 gms of phosphorus to react

with excess oxygen, which forms

tetraphosphorous dioxide P4O10, in 50 %

yield. In sept II reaction 25 % yield of

H3PO4 is obtained.

(ii) In order to remove organic sulphur from

coal, following reactions occur -

X–S–Y+ 2 NaOH X–O–Y+Na2S + H2O

CaCO3 CaO + CO2

Na2S + CO2 + H2O Na2CO3+ H2S

CaO + H2O Ca(OH)2

Na2CO3 + Ca(OH)2 CaCO3 + 2NaOH

Q.29 Assume 100 % yield, then number of moles of

H3PO4 would be obtained -

(A) 1.25 (B) 10.0

(C) 5.0 (D) 2.5

Sol.[B] P4 + SO2 P4O10

1 mol 1 mol

124 g P4 Produce = (124 + 160)g P4O10

Assuming 100% yield

1 g P4 Produce = 124

284

310 g g P4 Produce = 124

284× 310 = 710 g

P4O10 + 6H2O 4 H3PO4



1 mole 4 mole

284g P4O10 produce = 98g H3PO4 × 4


498× 710

= 98

7.987

= 10.00 mole

Q.30 Actual number of mole of H3PO4 obtained in

experiment conditions is -

(A) 5 (B) 2.5

(C) 1.25 (D) 0.3125

Sol.[C] P4 + SO2 P4O10

124g P4 Produce = 284 P4O10


284

310g P4 Produce = 124

284× 310 = 709.98

But P4O10 50% yield so

100

50=

98.709

weightActual

Actual weight = 355g

P4H10 + 6 H2O 4 H3PO4

1 mole 4 mole

284 g P4O10 Produce = 4 × 98g H3PO4

1 g P4O10 Produce = 284

392

355 g P4O10 Produce =284

392× 355

= 422.90

(But is 25% yield)

So that

100

25=

20.422

w

w = 122.28

Mole = 98

98.122= 1.25 mole

Q.31 In processing of 200 kg of coal having 4%

sulphur content the weight of lime stone that

must be decomposed to provide enough

Ca(OH)2 to regenerate the NaOH used in the

original leaching step is -

(A) 6.25 kg (B) 12.5 kg

(C) 18.75 kg (D) 25 kg

Sol.[D] X – S – Y + 2NaOH X O Y + Na2S + H2O

250 500

Na2CO3 + Ca[OH]2 CaCO3 + 2NaOH

250 500

CaO + H2O Ca(OH)2

250 250

CaCO3 CaO + CO2

250 250

wt. of S = 200 × 103 =

100

4= 8 × 10

3 gm = 250

mole

mole of CaCO3 = 250

wt of CaCO3………. = 250 × 100 = 25000 gm = 25

kg


Cis platin, an anticancer agent used for the

treatment of solid tumors, is prepared by the

reaction of ammonia with potassium

tetrachloroplatinate.

K2PtCl4 + 2NH3 [Pt(NH3)2Cl2] 2KCl

Cis - platin

Assume that 10 gm of K2PtCl4 & 10 gm of NH3 are

allowed to react. (K = 39, Pt = 195, Cl = 35.5)

Q.32 Number of moles of K2PtCl4 consumed -

(A) 0.048 (B) 0.024

(C) 0.012 (D) 0.096

Sol.[B] K2PtCl4 is L.R. hence it will completely

consumed

Q.33 Number of moles of NH3 consumed -

(A) 0.048 (B) 0.024

(C) 0.096 (D) 0.192

Sol.[A] Moles of ammonia will consumed twice that of

K2PtCl4

Q.34 Number of moles of excess reactant which

remains unreacted is -

(A) 0.024 (B) 0.34

(C) 0.54 (D) 0.56

Sol.[C] 0.58 – 0.04 = 0.54

no.of moles of NH3 which is unreacted.




The following chart shows the height of a

secret metal iodide precipitate that was formed

in a series of 10 test tubes. Each test tube

contained 3.0 mL of a metal nitrate solution

(1.0 mol L–1

with respect to the metal ions).

Measured volumes of 1.0 mol L–1

potassium

iodide (KI) solution were added to each tube in

turn to form the precipitate.

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1 2 3 4 5 6 7 8 9 10

Volume of KI solution added, mL

(Also showing number of test tube)

Hei

ght

of

pre

cipit

ate

in c

m

Q.35 What is the formula of the metal iodide

according to this chart ?

(A) MI4 (B) MI

(C) MI2 (D) MI3

Sol.[C]

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1 2 3 4 5 6 7 8 9 10

Volume of KI solution added, mL

(Also showing number of test tube)

Hei

ght

of

pre

cipit

ate

in c

m

Let metal Iodide be MIx

From test-tube 1 to 5, MIx is increasing with

addition of KI M(NO3)x is in excess.

However from test tube 6th

& second MIx

becomes

constant All Mx+

has been consumed after

addition of 6 ml of KI.

In 6th test-tube M

x+ + xI

– MIx

moles 3 mole 6 mole 3mole

= x

6 mol x = 2

metal iodide = MI2

Q.36 If the height of precipitate in test tubes 4 and 6

are considered to be authentic one, which test

tubes are reporting wrong height of precipitate ?

(A) 1 and 2 (B) 2 and 3

(C) 2 and 5 (D) 5, 7, 8, 9 and 10

Sol.[C] Clearly 2 and 5

Q.37 The molarity of I– (aq) ion in the 9

th test tube is

(A) 1.0 M (B) 0.5 M

(C) 0.33 M (D) 0.25 M

Sol.[D] I– added = 9 m mole

I– consumed = 6 m mol

I– left = 3 mole

V = 3 + 9 = 12 ml

[I–] =

12

3 =

4

1

Q.38 The constancy in height of precipitate after

addition of 6.0 mL or higher volume of KI

solution indicates that -

(A) the solution has become saturated in KI

(B) the weighing limit of balance has been

exceeded

(C) the entire metal ion has been precipitated

out

(D) the information available are insufficient to

interpret this very observation

Sol.[C] Clearly

TARUN KUMAR ASIWAL (15 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,

EXERCISE # 1

Question

based on

Sub-Atomic particles and Dalton's

atomic theory

Q.1 Proton is -

(A) Nucleus of deuterium

(B) Ionised hydrogen molecule

(C) Ionised hydrogen atom

(D) An -particle

Sol.[C] Proton is ionised hydrogen atom l.e., H+.

Q.2 Which is not deflected by magnetic field -

(A) Neutron (B) Positron

(C) Proton (D) Electron

Sol.[A] Neutron is charge less entity

Q.3 According to Dalton’s atomic theory, an atom

can -

(A) Be created

(B) Be destroyed

(C) take part in a chemical reaction

(D) None of these

Sol.[C] According to this theory an atom can neither be

created nor destroyed, i.e, indivisible.

Q.4 Arrange -particle(), electron (e–), proton(p)

and neutron (n) in increasing order of their e/m

value (specific charge, consider magnitude only

not sign) -

(A) < e– < p < n (B) n < < p < e

–

(C) n < p < < e– (D) e

– < p < n <

Sol.[D] m of is highest in comparing to others and m

of electron is minimum in comparing to others.

Question

based on Rutherford’s Experiment

Q.5 Rutherford’s alpha particle scattering

experiment eventually led to the conclusion

that -

(A) mass and energy are related

(B) electrons occupy space around the nucleus

(C) neutrons are burried deep in the nucleus

(D) the point of impact with matter can be

precisely determined

Sol. (B) Electrons in an atom occupy the extra nuclear

region

Q.6 Number of -particles scattered by an angle

in Rutherford's experiment are -

(A) Directly proportional to K.E2

(B) Directly proportional to z4

(C) Inversely proportional to e4

(D) Inversely proportional to K.E2

Sol.[D] N() 2E.K

1

Inversely proportional to K.E2

Electromagnetic waves, hydrogen

spectra & concept of quantization

::

Question

based on

Q.7 Wavelength of radio waves is -

(A) < microwaves (B) > microwaves

(C) infrared waves (D) u.v. rays

Sol.[B] The order of wavelength Radio waves >

microwaves > I.R. > visible > u.v > -rays >

-rays > cosmic rays

Q.8 The line spectra of two elements are not

identical because -

(A) the elements do not have the same number

of neutrons

(B) they have different mass number

(C) their outermost electrons are at different

energy levels

(D) they have different valencies

Sol.[C] Because their outermost electrons are at

different energy levels.

Q.9 A certain radio station broadcasts on a

frequency of 980 kHz (kilohertz). What is the

wavelength of electromagnetic radiation

broadcast by the radio station ?

(A) 306 m (B) 3.06 m

(C) 30.6 m (D) 3060 m

Sol.[A] =

C

or =

C =

3

8

10980

103

= 306 m


Q.10 Calculate the wavelength of the spectral line

when the electron in the hydrogen atom

undergoes a transition from fourth energy level

to second energy level ?

(A) 4.86 nm (B) 486 nm

(C) 48.6 nm (D) 4860 nm

Sol.[B]

1 = 109677

16

1–

4

1 cm

–1 or

= 3109677

16

cm

= 48.6 × 10–6

cm

= 486 nm

Q.11 The wave number of the first line of Balmer

series of hydrogen is 15200 cm–1. The wave

number of the corresponding line of Li2+ ion is-

(A) 15200cm–1 (B) 60800 cm–1

(C) 76000 cm–1 (D) 136800 cm–1

Sol.[D] (Li2+

) = z2 × 15200 cm

–1

= 32 × 15200 cm

–1

9 × 15200 cm–1

136 800 cm–1

Q.12 The frequency of one of the lines in Paschen

series of a hydrogen atom is 2.34 × 1014Hz.

The higher orbit, n2, which produces this

transitions is -

(A) three (B) four

(C) six (D) five

Sol.[D] = RHC

22

21 n

1–

n

1

n1 = 3

Q.13 In hydrogen spectrum, the series of lines

appearing in ultra violet region of

electromagnetic spectrum are called -

(A) Lyman lines (B) Balmer lines

(C) Pfund lines (D) Brackett lines

Sol.[A] Lyman series ultra violet region

Q.14 Which of the following series of lines in the

atomic spectrum of hydrogen appear in the

visible region ?

(A) Lyman (B) Paschen

(C) Brackett (D) Balmer

Sol.[D] Balmer series appear in the visible region

Q.15 Which of the following is not correct according

to Planck's quantum theory ?

(A) Energy is emitted or absorbed

discontinuously

(B) Energy of a quantum is directly

proportional to its frequency

(C) A photon is also a quantum of light

(D) Energy less than a quantum can also be

emitted or absorbed

Sol.[D] E = nh n = 1,2,3 ………

Only integer quantum can be emitted or

absorbed.

Q.16 To which electronic transition between Bohr

orbits in hydrogen, the second line in the

Balmer series belongs ?

(A) 3 2 (B) 4 2

(C) 5 2 (D) 6 2

Sol.[B] For Balmer series

n1 = 2 ; n2 = 3, 4, 5 ……….

for second line

n2 = 4

so 4 2

Question

based on Bohr’s atomic model

Q.17 The ratio of the radii of first three Bohr orbits is

(A) 1 : 05 : 3 (B) 1 : 2 : 3

(C) 1 : 4 : 9 (D) 1 : 8 : 27

Sol.[C] r n2

n = 1,2,3 …..

so ratio = 1 : 4 : 9

Q.18 The ionization energy of per mole of hydrogen

atom in terms of Rydberg constant (RH) is

given by the expression -

(A) RHhc (B) RHc

(C) 2 RH hc (D) RH NA hc

Q.19 The frequency of first line of Balmer series in

hydrogen atom is 0. The frequency of

corresponding line emitted by singly ionised

helium atom is -


(A) 2v0 (B) 4v0

(C) v0/2 (D) v0/4

Sol.[B] v = RH CZ2

22

21 n

1–

n

1

For first line of Balmer series

n1 = 2 ; n2 = 3 & Z = 1

0 = RHC

9

1–

4

1

0 = 36

5RHC ……….. (1)

For He+ atom Z = 2

= 4 × 36

5 RHC

or = 4 0

Q.20 Energy of third orbit of Bohr’s atom is -

(A) – 13.6 eV (B) – 3.4 eV

(C) – 1.51 eV (D) None of the three

Sol.[C] E = –13.6 2

2

n

z ev/atom

n = 3 & Z = 1

so, E = – 9

6.13 = –1.51 ev/atom

Q.21 If the radius of first Bohr orbit be a0, then the

radius of the third orbit would be -

(A) 3 × a0 (B) 6 × a0

(C) 9 × a0 (D) 1/9 × a0

Sol.[C] r = 0.529 Z

n 2

Å

n = 1, Z = 1

so r = 0.529 Å = 90

Therefore for third orbit

r = 0.529 × 9

or r = 9 × 90

Q.22 In H–atom electron jumps from 3rd to 2nd

energy level, the energy released is -

(A) 3.03 × 10–19

J/atom

(B) 1.03 × 10–19

J/atom

(C) 3.03 × 10–12

J/atom

(D) 6.06 × 10–19

J/atom

Sol.[A] E = 21 × 10–19

Z2

22

21 n

1–

n

1 J/atom

n1 = 2 & n2 = 3

Z = 1

So E = 21.8 × 10–19

× 12

9

1–

4

1

= 21.8 × 10–19

× 36

5 = 3.03 × 10

–19 J/atom

Q.23 The ratio of ionization energy of H and Be+3

is-

(A) 1 : 1 (B) 1 : 3

(C) 1 : 9 (D) 1 : 16

Sol.[D] IE = E– En

IE Z2

So, 3Be

H

IE

IE

= 16

1

ratio is 1 : 16

Q.24 The ionization energy of hydrogen atom (in the

ground state) is x kJ. The energy required for

an electron to jump from 2nd orbit to the 3rd

orbit will be -

(A) x/6 (B) 5x

(C) 7.2x (D) 5x/36

Sol.[D] IE = 1312 KJ/mole = x KJ

Energy required

E = 1312

22 3

1–

2

1

= 1312

9

1–

4

1

= 1312 × 36

5

or E = 36

x5 KJ/mole

Q.25 In two H atoms X and Y the electrons move

around the nucleus in circular orbits of radius r

and 4r respectively. The ratio of the times taken

by them to complete one revolution is -

(A) 1 : 4 (B) 1 : 2

(C) 1 : 8 (D) 2 : 1


Sol.[C] t = velocity

cetandis =

v

r2

or t n3

h

1v

nr 2

ty

tx =

3

3

2

1 =

8

1

Photoelectric effect, Dual Nature of

electron & Heisen berg’s uncertainty

principle

Question

based on

Q.26 If threshold wavelength (0) for ejection of

electron from metal is 330 nm, then work

function for the photoelectric emission is -

(A) 1.2 × 10–18

J (B) 1.2 × 10–20

J

(C) 6 × 10–19

J (D) 6 × 10–12

J

Sol.[C] W = h 0

C

W = 9

834–

10330

103106.6

= 33

1036.6 18

= 0.6 × 10–18

= 6 × 10–19

J

Q.27 The kinetic energy of the electron emitted when

light of frequency 3.5 × 1015

Hz falls on a metal

surface having threshold frequency 1.5 × 1015

Hz is (h = 6.6 × 10–34

Js)

(A) 1.32 × 10–18

J (B) 3.3 × 10–18

J

(C) 6.6 × 10–19

J (D) 1.98 × 10–19

J

Sol.[A] K.E. = h (v – v0)

= 6.6 × 10–34

(3.5 × 1015

– 1.5 × 1015

)

= 1.32 × 10–18

J

Q.28 Light of wavelength shines on a metal

surface with intensity x and the metal emits y

electrons per second of average energy, z. What

will happen to y and z if x is doubled ?

(A) y will be doubled and z will become half

(B) y will remain same and z will be doubled

(C) both y and z will be doubled

(D) y will be doubled but z will remain same

Sol.[D] No. of electrons emits Intensity

Energy Frequency

Q.29 A 200g cricket ball is thrown with a speed of

3.0 × 103 cm sec–1. What will be its de Broglie’s

wavelength ? [h = 6.6 × 10–27 g cm2 sec–1]

(A) 1.1 × 10–32 cm (B) 2.2 × 10–32 cm

(C) 0.55 × 10–32 cm (D) 11.0 × 10–32 cm

Sol.[A] = mv

h

= 3

27–

103200

106.6

= 6

106.6 32–

= 1.1 × 10–32

cm

Q.30 If uncertainty in the position of an electron is

zero, the uncertainty in its momentum would be

(A) zero (B) < h/(4)

(C) > h/(4) (D) infinite

Sol.[D] x × p 4

h

If x = 0

then p =

Q.31 Heisenberg uncertainty principle states that -

(A) Moving bodies exhibit both particle and

wave character

(B) Neither the position nor the momentum of

a particle can be precisely determined

(C) Simultaneous determination of position

and momentum of a microscopic particle is

not possible.

(D) Moving charged particles resemble electro-

magnetic waves in their behavior

Sol.[C] Simultaneous determination of position and

momentum (or velocity) of a microscopic

particle is not possible

Q.32 Calculate the uncertainty in velocity of a

cricket ball of mass 150 g if the uncertainty in

its position is 1 Å (h = 6.6 × 10–34 kg m2s–1)

(A) 3.5 × 10–24

ms–1

(B) 4.5 × 10–24

ms–1

(C) 3.5 × 10–24

cms–1

(D) 4.5 × 10–24

cms–1


Sol.[A] x × m.v 4

h

m = 150 g = 0.15 kg

x = 1Å = 10–10

metre

h = 6.6 × 10–34

kg m2 s

–1

= 3.14

V = 14.341015.0

106.610–

34–

= 3.5 × 10–24

ms–1

Schrodinger wave theory, quantum

number & shape of orbitals

Question

based on

Q.33 Which of the following statements is incorrect?

(A) Probabilities are found by solving

Schrodinger wave equation

(B) Energy of the electron in an atom at infinite

distance is zero and yet it is maximum

(C) Some spectral lines of an element may have

the same wave number

(D) The position and momentum of a rolling

ball can be measured accurately

Sol.[C] No two spectral lines of one element have same

wave number.

Q.34 For s-orbitals, since ( orbital wave function)

is independent of angles, the probability density

(2) is -

(A) also independent of angles

(B) spherically symmetric

(C) both (A) and (B) are correct

(D) both (A) and (B) are incorrect

Sol.[C] For spherical orbital

= f(r) (r = distance from nucleus)

since (orbital) is independent

of angles therefore probability (2)

also independent of angles and spherically

symmetric

Q.35 With the increasing principal quantum number,

the energy difference between adjacent energy

levels in H-atom -

(A) decreases

(B) increases

(C) remains constant

(D) decreases for low value of Z and increases

for higher value of Z

Sol.[A] E 2n

1

Q.36 How many electrons can fit into the orbitals

that comprise the 3rd

quantum shell n = 3 ?

(A) 2 (B) 8

(C) 18 (D) 32

Sol.[C] n = 3

electrons in 3rd

shell = 2n2

= 2 × 32 = 18

Q.37 Which of the following statements concerning

the four quantum numbers is false -

(A) n gives idea of the size of an orbital

(B) l gives the shape of an orbital

(C) ms gives the energy of the electron in the

orbital in absence of magnetic field

(D) ms gives the direction of spin angular

momentum of the electron in an orbital

Sol.[C] Magnetic quantum number (m) gives idea

about orientation of sub shells.

Q.38 Which of the following statements is not correct ?

(A) The shape of an atomic orbital depends on

the azimuthal quantum number

(B) The orientation of an atomic is given by

magnetic quantum number

(C) The energy of an electron in an atomic

orbital of multi electron atom depends on

the principal quantum number only

(D) The number of degenerate atomic orbitals

of one type depends on the values of

azimuthal and magnetic quantum numbers

Sol.[C] For multi electron atom energy depends on

n and (n + rule)

Question

based on Aufbau rule and e¯ configuration

Q.39 The manganese (Z = 25) has the outer configuration

(A)

d3s4

d3s4


(B)

(C)

d3s4

d3s4

(D)

Sol.[B] According to Aufbau principle atomic orbitals are filled in order of increasing energies.

Q.40 If the electronic structure of oxygen atom is

written as 1s2, 2s2

p2

it would

violate -

(A) Hund’s rule

(B) Pauli’s exclusion principle

(C) Both Hund’s and Pauli’s principles

(D) None of these

Sol.[A] According to this rule pairing of electrons does

not take place untill all the orbitals of the sub-

shell are singly occupied.

Q.41 A given orbital is labelled by the magnetic

quantum number m = –1. This can not be -

(A) s-orbital (B) d-orbital

(C) p-orbital (D) f-orbital

Sol.[A] For s-orbital value of m = 0 only.

True or false type questions

Q.42 The charge to mass ratio of the particles in

cathode rays is greater than that of the particles

in anode rays.

Sol. e/m ratio of e– > e/m ratio of P

Q.43 All the quantum numbers are not obtained from

Schrodinger wave equation.

Sol. Spin quantum no. can not determined by.

schrodinger wave equation

Q.44 The energy of the electron in 3d orbitals is less

than that in 4s orbitals in the hydrogen atom.

Sol. 3denergy > 4s, from (n + ) rule.

Q.45 A 22 yxd3

orbital has two angular nodes.

Sol. (d = 2), No. of nodal planes = 2 because = 2

Fill in the blanks type questions

Q.46 The 2px,2py and 2pz orbitals have same shape

but differ in their ..........

Sol. Px, P4, Pz are located perpendicular to each

other.

Q.47 Wave function of electrons in atoms and

molecules are called..........

Sol. 2 is called probable density known as orbital.

Q.48 The light radiations with discrete quantities of

energy are called ..........

Sol. Discrete quantity of energy is called photon.

Q.49 The wave character of the electrons was

experimentally verified by ..........

Sol. Devisson and germar experiment

Q.50 The kinetic energy of the electron in first orbit

of hydrogen is x kJ its kinetic energy in the

third orbit would be………

Sol. x = – 13.6 × 2

2

n

Z

x = – 13.6 × 1

1

E = – 13.6 × 2

2

n

Z = x ×

9

1

9

xE

EXERCISE # 2


Only single correct answer type

questions Part-A

Q.1 The quantum number not obtained from the

Schrodinger’s wave equation is -

(A) n (B)

(C) m (D) s

Sol.[D] Schrodinger's wave equation not releted to the

direction of e– so fourth quantum no. s not

obtained.

Q.2 The set of quantum numbers not applicable for

an electron in an atom is -

(A) n = 1, = 1, m = 1, s = + 1/2

(B) n = 1, = 0, m = 0, s = + 1/2

(C) n = 1, = 0, m = 0, s = – 1/2

(D) n = 2, = 0, m = 0, s = + 1/2

Sol.[A] Because n ≠

n = 1, = 1, m = 1, s = + ½ is not possible

Q.3 Maximum numbers of electrons in a subshell is

given by -

(A) (2+1) (B) 2(2+1)

(C) (2+1)2 (D) 2(2+1)2

Sol.[B] 2(2 + 1)

Q.4 Which one of the following represents an

impossible arrangement ?

n m s

(A) 3 2 –2 ½

(B) 4 0 0 ½

(C) 3 2 –3 ½

(D) 5 3 0 ½

Sol. [C] Because n > , so m = – to +

Q.5 Which of the following statements about nodal

planes is/are not true -

(A) A plane on which there is zero probability

of finding an electron

(B) A plane on which there is maximum

probability that the electron will be found

(C) 2 is non zero at nodal plane

(D) None of these

Sol.[B] A plane on which there is maximum probability

that the electron will be found

Q.6 Which of the following elements is represented

by the electronic configuration ?

1s

2s

2p

(A) Nitrogen (B) Fluorine

(C) Oxygen (D) None of these

Sol.[D] 10Ne = 1s22s

22p

6

Q.7 For the energy levels in an atom which one of

the following statement is correct ?

(A) The 4s sub-energy level is at a higher

energy than the 3d sub-energy level

(B) The M-energy level can have maximum of

32 electrons

(C) The second principal energy level can have

four orbitals and contain a maximum of 8

electrons

(D) The 5th main energy level can have

maximum of 49 electrons

Sol.[C] for n = 2,

2s, 2p

1 + 3 = 4 orbitals

2 + 6 = 8 electrons

Its have 4 orbitals and contain 8 electrons

Q.8 The electronic configurations of Cr24 and

Cu29 are abnormal -

(A) Due to extra stability of exactly half filled

and exactly fully filled sub shells

(B) Because they belong to d-block

(C) Both the above

(D) None of the above

Sol.[A] Cr24 – 1s22s

22p

63s

23p

64s

23d

4

(original configuration)

but – d5 – it's half filled state and that is state

most stable than other state.

Cr24

– 1s22s

22p

63s

23p

64s

13d

5

As above full filled state that is most stable

than other state.


Cu – 1s22s

22p

63s

23p

64s

23d

9

but 1s22s

22p

63s

23p

64s

13d

10 (full filled)

Q.9 The below configuration is not correct as it

violates

s

p d

(A) Only Hund’s rule

(B) Only Pauli’s exclusion principle

(C) (n + l) rule

(D) (Hund + Pauli) rule

Sol.[B] Only Pauli’s exclusion principle

Q.10 Wavelength of the first line of Paschen Series

is - (R = 109700 cm–1)

(A) [18750 Å] (B) [2854 Å]

(C) [3452 Å] (D) [6243 Å]

Sol.[A] [18750 Å]

Q.11 The maximum probability of finding electron

in the dxy orbital is -

(A) Along the x-axis

(B) Along the y-axis

(C) At an angle of 45º from the x and y-axis

(D) At an angle of 90º from the x and y-axis

Sol.[C] Because maximum probability of finding

electron in dxy orbitals at an angle of 45º from x

and y axis.

Q.12 The nucleus of an atom is located at

x = y = z = 0. If the probability of finding an

s-orbital electron in a tiny volume around

x = a, y = z = 0 is 1 × 10–5, what is the

probability of finding the electron in the same

sized volume around x = z = 0, y = a ?

(A) 1 × 10–5 (B) 1 × 10–5 × a

(C) 1 × 10–5 × a2 (D) 1 × 10–5 × a–1

Sol.[A] Both the points (x = a, y = z = 0) say and

(x = z = 0, y = a) say, B, lie on the surface of a

sphere of radius 'a' centred at nucleus. Since

s-orbital is spherical in shape, all the points

on it’s surface will have the same probability

(i.e. 1 × 10–5

) of finding a s-orbital electron.

s-orbital

x A a

a

B

y

z

Nucleus

Q.13 If n and are respectively the principal and

azimuthal quantum numbers, then the

expression for calculating the total number of

electrons in any energy level is -

(A)

n

0

)12(2

(B) 1n

1

)12(2

(C) 1n

0

)12(2

(D) 1n

0

)12(2

Sol.[D] Since, varies from 0 to (n – 1) for nth

energy

level.

Q.14 A photon was absorbed by a hydrogen atom in

its ground state and the electron was promoted

to the fifth orbit. When the excited atom

returned to its ground state, visible quanta were

emitted when electron made transition -

(A) 5 2 (B) 2 1

(C) 3 1 (D) 4 1

Sol.[A] Visible range Balmer series i.e. n1 = 2

So, first quanta is from 5 2

The other quanta will be 2 1

5 → 2 (visible)

n = 5

n = 2

n = 1 2 → 1

Q.15 What is the change in the orbit radius when the

electron in the hydrogen atom (Bohr model)

undergoes the first Paschen transition ?

(A) 4.23 × 10–10 m (B) 0.35 × 10–10 m

(C) 3.7 × 10–10 m (D) 1.587 × 10–10 m

Sol.[C] r3 = 0.529 × Z

n 2

Å

r4 = 0.529 × Z

n 2

Å


r4 – r3 = 0.529 × 1

16 – 0.529 ×

1

9Å

= 0.529 × 10–10

[16 – 9] m

= 7 × 0.529 × 10–10

= 3.7 × 10–10

m

Q.16 In centre-symmetrical system, the orbital angular

momentum, a measure of the momentum of a

particle travelling around the nucleus, is

quantised. Its magnitude is -

(A) )1( 2

h

(B) )1( 2

h

(C) )1s(s 2

h

(D) )1s(s 2

h

Sol.[A] )1( 2

h

Q.17 Ultraviolet light of 6.2 eV falls on a aluminium

surface (work function = 4.2 eV). The kinetic

energy (in joule) of the fastest electron emitted

is approximately -

(A) 3 × 10–21 (B) 3 × 10–19

(C) 3 × 10–17 (D) 3 × 10–15

Sol.[B] h = h0 + KE

KE = h – h0 h = 6.2 eV

= 6.2 – 4.2 h0 = 4.2 eV

= 2.00 × 1.6 × 10–19

KE = 3.2 × 10–19

J

Q.18 An electron, a proton and an alpha particle have

kinetic energies of 16E, 4E and E respectively.

What is the qualitative order of their de-Broglie

wavelengths ?

(A) e > p = (B) p = > e

(C) p > e > (D) < e >> p

Sol.[A] KEm2

h =

31101.9E162

h

= e.

pe

Q.19 One energy difference between the states

n = 2 and n = 3 is E eV, in hydrogen atom. The

ionisation potential of H atom is -

(A) 3.2 E (B) 5. 6E

(C) 7.2 E (D) 13.2 E

Sol.[C] Ionization energy = E – En

= 0 –

eV6.13

n

Z2

2

= 2

2

n

Z× 13.6 eV

= 13.6 Z2

22

21 n

1

n

1

n1 = 2, n2 = 3

= 13.6 × 1

94

49

= 9

5× 13.6 = 7.2 E

Q.20 Magnetic moments of V(Z = 23), Cr(Z = 24),

Mn(Z = 25) are x, y, z. Hence -

(A) x = y = z (B) x < y < z

(C) x < z < y (D) z < y < x

Sol.[C] BM)2n(n

n = no. of unpaired electrons

Q.21 The speed of a proton is one hundredth of the

speed of light in vacuum. What is its de-Broglie

wavelength ? Assume that one mole of protons

has a mass equal to one gram [h = 6.626 ×

10–27 erg sec]

(A) 13.31 × 10–3 Å (B) 1.33 × 10–3 Å

(C) 13.13 × 10–2 Å (D) 1.31 × 10–2 Å

Sol.[B] = PmV

h

Vp = 100

Vc =100

103 8= 3 × 10

6

M = 1.67 × 10–27

kg

= 627

34

1031067.1

1062.6

= 1.33 × 10–13

m

= 1.33 × 10–3

Å


Q.22 If the shortest wavelength of H atom in Lyman

series is x, then longest wavelength in Balmer

series of He+ is -

(A) 5

x9 (B)

5

x36 (C)

4

x (C)

9

x5

Sol.[A] Shortest wavelength of H-atom

n1 = 1 n2 = Z = 1

1 = RH × 1 ×

1

1

1 E =

hc

1 = RH E = n so n2 =

1 = x

longest wavelength n1 = 2, n2 = 3

1 = RH × 4 ×

9

1

4

1

= RH × 4 ×

36

49 =

36

5

1 =

36

54R H

1 =

9

5× x

5

x9

Q.23 The specific charge of a proton is

9.6 × 107 C kg–1, then for an -particles it will

be -

(A) 2.4 × 107C kg–1 (B) 4.8 × 107C kg–1

(C) 19.2 × 107C kg–1 (D) 38.4 × 107C kg–1

Sol.[B] For Proton e/m = 9.6 × 107 C/kg

For particle = m4

e2=

2

1e/m

=2

1× 9.6 ×10

7

= 4.8 ×107 C/kg

Q.24 The dissociation energy of H2 is 430.53KJ mol–1

.

If H2 is dissociated by illuminating with the

radiation of wavelength 253.7 nm, the fraction

of the radiant energy which will be converted

into kinetic energy is given by -

(A) 8.76 % (B) 12.33 %

(C) 11.3 % (D) 100%

Sol.[A]

E = 430.53 kJ/ mol

E1= 23

3

1002.6

21053.430

J/atom

E2 =

hc

E2 = 9

834

107.253

103106.6

=1

2

E

E× 100

= 21053.430107.253

1002.6103106.639

23834

×100

= 9.425

10618

= 8.76%

Q.25 In an electron microscope, electrons are

accelerated to great velocities. Calculate the

wavelength of an electron travelling with a

velocity of 7.0 megameters per second. The

mass of an electron is 9.1 × 10–28g

(A) 1.0 × 10–13 m (B) 1.0 × 10–7 m

(C) 1.0 m (D) 1.0 × 10–10m

Sol.[D] Velocity = 7 × 106 meter

= mv

h=

631

34

107101.9

106.6

= 1×10–10

m

Q.26 What are the values of the orbital angular

momentum of an electron in the orbitals 1s, 3s,

3d and 2p ?

(A) 0, 0, 62

h

, 2

2

h

(B) 1, 1, 42

h

, 2

2

h

(C) 0, 1, 62

h

, 3

2

h

(D) 0, 0, 202

h

, 6

2

h

Sol.[A] From )1( 2

h

Values of l for 1s, 2s, 3d and 2p are 0, 0, 2, 1.

0, 0, 62

h

, 2

2

h


Q.27 The energy difference between two electronic

states is 46 .12 kcal /mole. What will be the

frequency of the light emitted when an electron

drops from the higher to the lower energy

state? (Planck' constant = 9.52 × 10–14 kcal sec

mole–1)

(A) 4.84 × 1015 cycles sec–1

(B) 4.84 × 10–5 cycles sec–1

(C) 4.84 × 10–12 cycles sec–1

(D) 4.84 × 1014 cycles sec–1

Sol.[D] E =

hc

E2 – E1 =

hc

348

3

106.6103

1012.46

=

=

c=

A834

38

N103106.6

1012.46103

= 4.84 ×1014

cyc/sec.

Q.28 The radii of two of the first four Bohr orbits of

the hydrogen atom are in the ratio 1 : 4. The

energy difference between them may be -

(A) Either 12.09 eV or 3.4 eV

(B) Either 2.55 eV or 10.2 eV

(C) Either 13.6 eV or 3.4 eV

(D) Either 3.4 eV or 0.85 eV

Sol.[B] Radius ratio of K and L shell = 1 : 4

Energy difference = – 3.4 – (–13.6) = 10.2 eV

Radius ratio of level L and N shell = 2 : 4

= – 0.85 – (– 3.4)

= 2.55 eV

Q.29 It is known that atoms contain protons,

neutrons and electrons. If the mass of neutron is

assumed to be half of its original value whereas

that of electron is assumed to be twice of this

original value. The atomic mass of 6C12 will be-

(A) Twice (B) 75% less

(C) 25% less (D) one-half of its

Sol.[C] A = P + n A = Z + n

A = 12

Z = 6

n = 12 – 6

n = 6

= 12

9×100 = 75%

= 100 – 75

= 25% less mass

Q.30 The Z-component of angular momentum of an

electron in an atomic orbital is governed by the -

(A) Principal quantum number

(B) Azimuthal quantum number

(C) Magnetic quantum number

(D) Spin quantum number

Sol.[C] me = )1( cos

Q.31 The angle made by angular momentum vector

of an electron with Z-axis is -

(A) cos = / m (B) cos = m

(C) cos = m

)1( (D) cos=

)1(

m

Sol.[D] cos = )1(

m

Q.32 In an atomic orbital the sign of the lobes

indicates the -

(A) Sign of the probability distribution

(B) Sign of charge

(C) Sign of wave function

(D) presence or absence of electron

Sol.[C] Theory

Q.33 Which of the following symbols represent an

atomic orbital ?

(A) n, , m = Rn m

(B) n, , m = Rn, m

(C) n, , m = Rn ,m m

(D) n, , m = Rn, , m m

Sol.[C] Polar form of Schrodinger equation.

One or more than one correct

answer type questions Part-B

Q.34 Which of the following properties is/are

proportional to the energy of the

electromagnetic radiation ?

(A) Frequency (B) Wave number


(C) Wavelength (D) Number of photons

Sol. [A, B, D]

E = c

h E , E

=

1,

E h

Q.35 Which of the following statements are

incorrect?

(A) There are five unpaired electrons in (n –1)d

suborbit in Fe3+

(B) Fe3+, Mn+ and Cr all having 24 electrons

will have same value of magnetic moment

(C) Copper (I) chloride is coloured salt

(D) Every coloured ion is paramagnetic

Sol.[B, C]

2727Cr

24125Mn

23326Fe3

saltcolourednot

isitsoeunpairedno

d3p3s3p2s2s1Cu 1062622

Q.36 If Aufbau rule and Hund’s rule are not

considered, which of the following statements

are correct ?

(A) Fe2+ will have configuration as

|Ar|

(B) Cu2+ is colourless ion

(C) Magnetic moment of the Mn is 3 BM

(D) K+ is of d-block

Sol.[A,C] In option (A) is not considered hunds

rule and in option (C) Mn has 3 unpaired

electron so 3 BM

Q.37 Which of the following orbitals have no

spherical nodes ?

(A) 1s (B) 2s (C) 2p (D) 3p

Sol.[A,C]

Spherical nodes = n – – 1

Q.38 In which of the following sets of

orbitals, electrons have equal orbital angular

momentum ?

(A) 1s and 2s (B) 2s and 2p

(C) 2p and 3p (D) 3p and 3d

Sol.[A, C]

)1( 2

h = angular momentum

Q.39 Which of the following sets of quantum

number are correct ?

(A) n = 3, l = 2, m = + 1, s = +2

1

(B) n = 3, l = 3, m = + 3, s = +2

1

(C) n = 4, l = 0, m = 0, s = – 2

1

(D) n = 5, l = 2, m = + 4, s = – 2

1

Sol.[A, B] 3d, 4s

Q.40 Rutherford’s experiment established that :

(A) Inside the atom there is a heavy positive

centre

(B) Nucleus contains protons and neutrons

(C) Most of the space in the atoms is empty

(D) Size of the nucleus is very small

Sol.[A, C, D]

Rutherford had not discovered about neutrons

Q.41 Which of the following statements are

incorrect ?

(A) For designating orbitals three quantum

numbers are needed

(B) The second ionization energy of helium is

4 times, the first ionization of hydrogen

(C) The third ionization energy of lithium is 9

times, the first ionization of hydrogen

(D) Radius of third orbit of Li2+ is 3 times the

radius of third orbit of hydrogen atom

Sol.[C, D]

(IE)Z = 2

2H

n

Z.)E.I(

(IE)Li = 2

2H

3

1.)E.I(

HLi .)E.I(.)E.I(9

Q.42 Which of the following statements (regarding

an atom of H) are correct ?

(A) Kinetic energy of the electron is maximum

in the first orbit

(B) Potential energy of the electron is maximum in the first orbit

(C) Radius of the second orbit is four times the

radius of the first orbit


(D) Various energy levels are equally spaced

Sol.[A, C]

(A) E = – 13.6 × 2

2

n

Z n E

(B) PE = + 13.6 × 2

2

n

Z n E

(C) r2 = Z

)2( 2

, r1 = Z

)1( 2

1

2

r

r=

1

4

r2 = 4r1

(D) Different

Q.43 Which of the following transition in H–atom

would result in emission of radiations of same

frequency ?

(A) 4s 3p (B) 4d 3p

(C) 5s 4s (D) 3s 2p

Sol.[A, B]

Because frequency depends only shell no. not

depends orbitals.

Assertion-Reason type questions Part-C

Choose any one of the following four

responses.

(A) If both Assertion and Reason are true

and the Reason is correct explanation of

the Assertion.

(B) If both Assertion and Reason are true

but the Reason is not a correct

explanation of the Assertion.

(C) If Assertion is true but the Reason is

false.

(D) If Assertion is false but Reason is true.

Q.44 Assertion : The charge to mass ratio of the

particles in anode rays depends on nature of the

gas taken in the discharge tube.

Reason : The particles in anode rays carry

positive charge.

Sol. [B] anode rays deflect to –ve terminals.

Q.45 Assertion : s-orbital cannot accommodate more

than two electrons.

Reason : s-orbitals are spherically symmetrical.

Sol. [B] due to terminals.

Q.46 Assertion : Kinetic energy of photoelectrons is

directly proportional to the intensity of the incident radiation

Reason : Each photon of light causes the

emission of only one photo electron.

Sol. [D] Intensity is not directly proparhance to K.E

Q.47 Assertion : The existence of three unpaired

electrons in phosphorous atom can be

explained on the basis of Hund’s rule.

Reason : According to Hund’s rule, the

degenerate orbitals are first singly occupied and

only then pairing takes place.

Sol. [A] See hunds statements

Column Matching Part-D

Q.48

Column-A Column-B

(A) Shape of the orbital (i) Principle

quantum number

(B) Size of the orbital (ii) Spin quantum

number

(C) Spatial orientation

of the orbital

(iii) Angular

momentum

quantum number

(D) Spin angular

momentum

(iv) Magnetic

quantum number

Sol. (a) (iii), (b) (i), (c) (iv), (d) (ii)

Fact

Q.49

Column-A Column-B

(A) Orbitals having equal

energy

(i) 3p, 3d

(B) Orbitals having zero

orbital angular momentum

(ii) 2s and 3s

(C) Orbitals with only one

spherical node

(iii) Degenerate

orbitals

(D) Orbitals having

directional character

(iv) 2s and 3p

Sol. (a) (iii) fact

(b) (ii) = 0

(c) (iv) n – – 1 (d) (i) fact

Q.50 Match the column :

Column-A Column-B

(A) R

r

(i) 3pz


(B)

4

r2R

2

r

(ii) 4py

(C) Angular probability is dependent on

and

(iii) 3s

(D) At least one angular node is present (iv) 3px

Sol. (A) 2 radial nodes from graph & s-orbital

3s

(B) 2 radial nodes 4py & 3s

(C) Angle depedent probability 4py, 3px

(D) Angular node 3p2, 4py, 3px

EXERCISE # 3

Subjective Type Questions Part-A

Q.1 The dye acriflavine, when dissolved in water,

has its maximum light absorption at 4530 Å

and its maximum fluorescence emission at

5080 Å. The number of fluorescene quanta is,

on the average, 53% of the number of quanta

absorbed. Using the wavelength of maximum

absorption and emission, what percentage of

absorbed energy emitted of fluorescence ?

Sol. Energy – E1

1 = 5080Å, 2 = 4530Å

E1 = h =

c

= 21

hc

E1 = 10

834

10)45305080(

103106.6

= 0.36 × 10–37

E2 = h = )(

hc

21

1 = 100

534530 = 453 × 5.3 Å = 2400Å

2 = 100

475080 = 4.7 × 508 Å = 2387Å

E2 = )(

hc

21

=)23872400(

103106.6 834

= 16.60 × 10–37

% 100 ×

1

2

E

E=

36.0

6.16×100 = 47.28 %

Q.2 The line at 434 nm in the Balmer series of the

hydrogen spectrum corresponds to a transition

of an electron from the nth to second Bohr orbit.

What is the value of n ?

Sol. = 434 × 10–9

× 1010

Å = 43.4 Å

1= RZ

2

22

21 n

1

n

1

n1 = 2, n2 = n Solving that equation we can get n = 5

Q.3 When would the wavelength associated with an

electron be equal to wavelength of proton?

(mass of e= 9 × 10–28 g ;

mass of proton = 1.6725 × 10–24 g)

Sol. e = p

ee vm

h=

ppvm

h

meve = mpvp

p

e

v

v=

e

p

m

m=

28

24

109

1067.1

= 1.85 × 10

3

Q.4 Point out the angular momentum of an electron in

(a) 4s orbital (b) 3p orbital

(c) 4th orbit

Sol. Angular momentum for subshell 4s, 3p and 4th orbit

(A) for S, = 0 )1( .2

h= 0

(B) for P , = 1 )1( .2

h

= 2 .2

h=

2

h

(C) For shell 4th, n = 4


=

nh =

h4 =

h2

Q.5 (a) The wave number of the first line in the

Balmer series of Be3+ is 2.43 × 105 cm–1. What

is the wave number of the second line of the

Paschen series of Li2+ ?

(b) In ions like He+, Li2+, Be3+ how and why

does the value of the Rydberg constant

vary ?

Sol. (A) For Be3+

Z = 4 For first line (Balmer series) n1 = 2, n2 = 3 = 2.43 ×10

5 m

–1

= RH (4)2

22 3

1

2

1 =

9

20 …….(1)

For Li2+

, Z = 3 for second line (Paschen series)

n1 = 3 and n2 = 5

= RH(3)2

22 5

1

3

1 =

25

16 …….(2)

dividing (1)/(2)

51043.2 =

9

20×

10

25 = 0.70 × 10

5

= 0.7 × 106 m

–1

(B) From formula

RH = 2Z

1

)nn(

nn21

22

22

21

RH depend on Z, n1 and n2 so increase Z, n1, n2 as RH vary with them

Q.6 Calculate the wavelength in Å of the photon

that is emitted when an electron in Bohr orbit

with n = 2 returns to orbit with n = 1 in H atom.

The ionisation potential of the ground state of

H-atom is 2.17 × 10–11 erg.

Sol.

1= RHZ

2

22

21 n

1

n

1

n1 = 1 n2 = 2

= RH

4

11

HR

1= 912Å

1 = RH ×

4

3

= HR3

4=

3

4× 912 = 1216 Å

Q.7 Two particles A and B are in motion. If the

wavelength associated with the particle A is

5 × 10–8 m, calculate the wavelength of particle

B if its momentum is half of A.

Sol. mvr = 2

nh =

p

h

h

mv =

r2

n

…(1) =

mv

h

1 =

r2

n

r2

n

=

8105

1

…(2)

Momentum of B is half of A

2

mvr =

2

nh

h2

mv=

r2

n

From (2)

1 =

8105

1

×2

1

= 10 × 10–8

m

Q.8 The first ionization energy of H is 21.79 × 10–19 J.

Determine the second ionization energy of He

atom.

Sol. I.P. = 13.6 × 2

2

n

Z = IP of H ×

2

2

n

Z

He = 21.79 × 10–19

× 2

2

1

Z

= 87.16 × 10–19

Joule

Q.9 How many times larger is the spacing between

the energy levels with n = 3 and n = 8 spacing

between the energy level with n = 8 and n = 9

for the hydrogen atom ?

Sol. r1 = 0.529

z

n

z

n 22

21

= 0.529 × [82 – 3

2]

= 29.095 Å

r2 = 0.529 × (92 – 8

2)

= 8.9 Å

2

1

r

r=

9.8

095.29 = 3.23 (spacing)

Q.10 To what series does the spectral lines of atomic

hydrogen belong if its wave number is equal to

the difference between the wave numbers of the

following two lines of the Balmer series : 486.1

and 410.2 nm ? What is the wavelength of that

line ?


Sol. Given 1 = 486.1 × 10–9

m

2 = 410.2 × 10–9

m

12 = 2

1

–

1

1

= RH

22

2 n

1

2

1 – RH

21

2 n

1

2

1

= RH

22

21 n

1

n

1

For I case of Balmer series

1

1

= RH

21

2 n

1

2

1= 109678

21

2 n

1

2

1

n1 = 4

For II case

2

1

=

7102.410

1

= 109678

22

2 n

1

2

1

n2 = 6

Transition 6 to 4

=

1=109678

22 6

1

4

1= 2.63 × 10

–4 cm

Q.11 Calculate the minimum uncertainity in velocity

of a particle of mass 1.1 × 10–27 kg if uncertainity

in its position is 3 × 10–10 cm.

Sol. v.x m4

h

v m.4.x

h

= 27210

34

101.114.34m10103

106.6

= 1.5 × 10–4

m

Q.12 Calculate the number of photons emitted in

10 hours by a 60 W sodium lamp.

(photon = 5893 Å).

Sol. E =

nhc E = 60 watt by 10 hours

= 60 × 10 × 60 × 60 Joule/sec

= 216 × 104 J/sec.

n = hc

Exd

= 834

104

103106.6

10589310216

= 6.4 × 1024

meter

Q.13 Calculate total spin, magnetic moment for the

atoms having atomic number 7, 24, 34 and 36.

Sol. Z7 =

1s2 2s

2 2p

3

0

0 ±½ ±½ ±½

= 2

3

as it for 24, 34 and 36 z24

= 3

z34

= 1

z36

= 0

Magnetic moment

For 7 = )2n(n = BM

= )23(3 = 15 BM

So on for 24, 34 and 36 = 48 , 8 , 0 BM

Q.14 Magnetic moment of X3+ ion of 3d series is

35 B.M. What is atomic number of X3+ ?

Sol. MB = )2n(n = 35

So n = 5

X3+

= 5 + 3 = 8

= 3d63s

2

So Fe = 26

Q.15 Two hydrogen atoms collide head on and end

up with zero kinetic energy. Each then emits a

photon with a wavelength 121.6 nm. Which

transition leads to this wavelength ? How fast

the hydrogen atoms were travelling before the

collision ? (Given : RH = 1.097 × 107 m–1 and

mH = 1.67 × 10–27 kg)

Sol. Wave length in UV region and thus n1 = 1

1 = RH

22 n

1

1

1

9106.121

1

= 1.097 × 107

22 n

1

1

1

n = 2

Also the energy released is due to collision

and all the kinetic energy is released in form of

photon

2

1mv

2 =

hc

2

1× 1.67 × 10

–27 × v

2 =

9–

834

106.121

1031062.6

v = 4.43 × 104 m/sec

Q.16 When a certain metal was irradiated with light

of frequency 3.2 × 1016 Hz, the photoelectrons


emitted had twice the kinetic energy as did

photoelectrons emitted when the same metal

was irradiated with light of frequency

2.0 × 1016 Hz. Calculate v0 for the metal.

Sol. 2maxmv

2

2= h(3.2×10

16–0) …(1)

2maxmv

2

1= h(2.0×10

16–0) …(2)

Solving equation (1) and (2)

1

2=

016

016

100.2

102.3

0.8 × 1016

Hz

8 × 1015

Hz

Q.17 Calculate the circumference of the 4th Bohr

orbit for an electron travelling with a velocity

of 2.19×106m/s.

Sol. mvr = 2

nh

2r = mv

nh=

631

34

1019.2101.9

106.64

2r = 13.3 ×10–10

meter

Q.18 1.53 g of hydrogen is excited by irradiation. At

a certain instant, 10% of the atoms are at the

excited level of energy –328 kJ mol–1 and 2%

of the atoms are at the excited level of energy –

146 kJ mole–1. The remaining atoms are in the

ground state. Calculate how much enegy will

be evolved when all the excited atoms return to

the ground state.

Sol. mole of H = 1.53

E = – 1300.96 2

2

n

z KJ/mol

E = 1.53 × 0.1 × (1300.96 – 328)

+ 1.53 × 0.02 (300.96 – 146)

= 184.2 KJ

Q.19 The photochemical dissociation of oxygen

results in the production of two oxygen atoms, one

in the ground state and one in the excited state

O2 h

O + O*. The maximum wavelength ,

needed for this is 174 nm. If the excitation energy

O O• is 3.15 × 10–19 J how much energy in

kJ/mol is needed for the dissociation of one of

oxygen into normal (i.e. ground state) atoms ?

Sol. O2

E O + O

O + O

3.15 × 10–19

J./e

Given : 0 – 0 energy = 3.15 × 10–19

J/al

and O2 0 + 0 energy required = E

i.e. E = h0 =

hc

= 9–

834–

10174

1031062.6

J

= 0.114 × 10–34 + 8 + 9

0.114 × 10 –34+17

= 0.114 × 10–17

J.

11.4 × 10–19

J.

Energy required for origin process

Erequired = E – 3.15 × 10–19

J

= 11.41 × 10–19

– 3.15 × 10–19

= 8.34 × 10–19

J/atm

For per mole.

8.34 × 10–19

× 6 × 1023

50 100

23 × 10

2

a 1000

105023 2 = 502.3 KJ/mole

so almost. covert

Q.20 Electrons of energy 12.1 eV are fired at the

hydrogen atom in a gas discharge tube.

Determine the wavelength of the lines that can

be emitted by hydrogen.

Sol. 1212 Å, 1022Å, 5545Å

Q.21 Calculate the angular frequency ( = v/r) of an

electron occupying the second Bohr orbit of

He+ ion.

Sol. Velocity of an e– in He

+ ion in an orbit

= nh

Ze2 2 …(1)

Radius of He+ ion in an orbit

= Zme4

hn22

22

…(2)

By Equ. (1) and Equ. (2)

w = r

u =

32

423

hn

meZ8

= 3273

4102823

)106.6()2(

)108.4()101.9()2()14.3(8

= 2.067 × 1016

sec–1

Q.22 As the speed of a particle of mass, m, increases,

its de-Broglie wavelength decreases. What is

the rate of change of with the kinetic energy,


E for an electron with = 1 Å ? Mass of the

electron = 0.91 × 10–30 kg.

Sol. = Em2

h

2 =

m

2

E2

h

dE

d =

M2

h 2

dE

d =

2

2

mE4

h–

dE

d = –

10217–31–

234–

10)104.2(1.94

)1062.6(

= 2.09 × 106 m/J

mu

h =

22

2

4m

h

2

mE2

h2

= 2

E = 2

2

m2

h

3–

234–

101.92

)1062.6(

× (10

–10)2

= 2.4 × 10–17

J

Q.23 A single electron atom has nuclear charge +Ze

where Z is atomic number and e is electronic

charge. It requires 47.2 eV to excite the

electron from the second Bohr orbit to third

Bohr orbit. Find.

(a) The atomic number of element.

(b) The energy required for transition of

electron from third to fourth orbit.

(c) The wavelength required to remove

electron from first Bohr orbit to infinity.

(d) The kinetic energy of electron in first Bohr

orbit.

Sol. (A) 1eV = 1.6 × 10–12

erg

E =

hc = E3 – E2 = RHZ

2

22 3

1

2

1

z2 = 22.6

z 5

(B) E = E4 – E3 = RHZ2

22 4

1

3

1

= 23.9 × 10–12

erg = 23.9 × 10–19

J

(C) v =

1 = RHZ

2

22

1

1

1

= 3.65 × 10–7

cm = 36.5 Å

(D) K.E. = 2

1mv

2 =

2

1m × (2.188 × 10

8)

2

= 5.45 × 10–10

erg

Q.24 The de Broglie wavelength of electron of He+

ion is 3.329 Å. If the photon emitted upon de-

excitation of this He+ ion is made to hit H-atom

in its ground state so as to liberate electron

from it, what will be the de-Broglie's

wavelength of photoelectron.

Q.25 The subshell that arises after f is called g

subshell.

(a) How many g orbitals are present in the g

subshell ?

(b) In what principal electronic shell would the

g subshell first occur and what is the total

number of orbitals in this principal shell?

Sol. (A) g

orbitals = 9

(B) (i) g start in n = 5

(ii) n2 = (5)

2 = 25 orbitals in principal shell

Q.26 What is the speed of an electron whose de

Broglie’s wavelength is 1 nm ?

(A) 7.36 × 106 m/s (B) 7.36 × 10

5 m/s

(C) 1.35 × 106 m/s (D) 4.821 × 10

6 m/s

Sol.[B] d = mv

h

V = m

h

d =

)101.9()101(

106.6319

34

V = 7.36 × 105 m/s

Q.27 Quantum efficiency, is defined as-

= absorbedphotonof.No

reactedmoleculesof.No

A glass vessel containing large quantities of H2

(g) & Cl2 (g) is irradiated with a wavelength of

500 nm with a quantum efficiency equal to 104.

The entire vessel is absorbed 10 J of energy.


After the irradiation the gas mixture remaining

in the vessel is washed with 100 L of water

thoroughly. What will be the pH of the

washwater in nearest possible integer ?

(Given : NA = 6 × 1023

; c = 3 × 108 m/s ;

h = 6.625 × 10–34

J s)

Sol. 10 J energy no. of photon = photonofE

10

N =

hc

10 =

834

9

103106.6

1050010

= 2.5 ×1019

= 10–4

No. of molecules reacted

= 2.5 × 1023

H2(g) + C2(g) 2HCl(g)

nHCl produce = 2 ×

0

5.2 × 10

23 ×

AN

1

= 0.83 moles

pH = –log

100

83.0

pH = 2 – log 0.83 = 2.06

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