7/29/2019 Ans Mole

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Answer for Mole concept and Volumetric analysis

1. Number of moles of magnesium present= 0.450/24.31=0.0185Number of moles of nitrogen present = (0.623-0.450)/ 14.01 = 0.0123

Ratio of No of moles = Mg : N = 0.0185 : 0.0123

= 1.5/1 = 3:2

The empirical formula is Mg3N2

2. a) Mass of C in vitamin C = 0.2998 x 12/44.01 = 0.0818gMass of H in vitamin C = 0.0819 x 2/18.02 = 0.0092g

Mass of O in vitamin C = 0.2-0.0092-0.0818 = 0.109g

Relative mole ratio of atoms:

C : H : O = 1 : 1.33 : 1 = 3 : 4 : 3

Empirical formula = C3H4O3

b) molecular mass of empirical formula = 88

88n = 180

n = 2.045

n is an integer, n=2

Thus, molecular formula of vitamin C is C6H8O6

3. 2LiOH + CO2 Li2CO3 + H2ONo of moles of LiOH in 1.00kg = 1000 / 23.95 = 41.81mole CO2 : 2mole LiOH

No of moles of CO2 = 41.8 /2 = 20.9 mol

Mass of CO2 absorbed = 20.9 x 44 = 920g

4. (a) 2Mg + O2 2MgONo of moles of Mg = 2.43 / 24.3 = 0.1 mol

Since oxygen is in excess, all Mg will be converted to MgO

Mass of MgO formed = 0.10X 40.3 = 4.03g

(b) No of mole of oxygen = 1.28 / 32 = 0.04 mol

2 mole of Mg : 1 mole of O2

0.04 mole of O2 requires 0.08 mole Mg . Since there are 0.1 mole of Mg,

thus Mg is in excess (Oxygen is limiting).

All O2 will be reacted.

No of mole of MgO formed = 2 x 0.04 = 0.08

Mass of MgO formed = 0.08 x 40.3 = 3.22g

5. a) (CH3)3COH + HCl (CH3)3CCl + H2O

7/29/2019 Ans Mole

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Answer for Mole concept and Volumetric analysis

b) No of moles of (CH3)3COH = 25/74 = 0.337

No of moles of HcL = 36/ 36.5 = 0.986

HCl is in excess , thus (CH3)3COH is limiting reagent.

c) (Experimental )No of moles of (CH3)3CCl = 28/92.5 = 0.303

% yield = Experimental / theoretical = 0.0303/ 0.337 X 100% = 89.9%

6. (a )2X(s ) + 6HCl(aq) 2XCl3(aq) + 3H2(g)X2O3(s) + 6HCl(aq) 2XCl3(aq) + 3H2O (l)

(b) In reaction I

No of mole of X= 0.5724/ 3 = 0.1908mol

Relative atomic mass = 16.5/ 0.1908 = 86.5

In reaction II

No of mole of X2O3 = 0.5724/6 = 0.954molLet the relative atomic mass of X be A , 16.5/ (2A+16x3) = 0.954

Relative atomic mass of X = 62.5

Thus the smallest relative atomic mass of X is 62.5 , the greatest relative

atomic mass of X is 86.5

(c) The only trivalent metal with relative atomic mass in the range of 62.5 to

86.5 is gallium (Ga).

7.

(a) The solution turns brown ( due to complex formation)I2 + I

- I3

-(aq)

(b) I2 is volatile, no accurate weighing of the mass of iodine solid.

(c) 78.5 cm3

(d) Starch indicator will form complex with I2 reversibly and turns blue black.

The end point can be observed easily.

(e) Near the end point. Starch indicator will combine irreversibly with iodine

when the concentration of iodine is high at start.

8 (a) Volume of CO2 produced = 100 cm3

Volume of unreacted O2 = 100 cm3

Volume of N2 = 250-100-100= 500 cm3

Using Avogadros law,

1/50 = x/100 = (y/2)/100

x = 2, y= 2

Formula is C2H2

(b) C2H2(g)+ 2O2 (g) 2CO2 (g) + N2 (g)

7/29/2019 Ans Mole

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Answer for Mole concept and Volumetric analysis

9. (a) Mg + H2SO4MgSO4 + H2(b) Phenolphthalein changes from colourless to pink

(accept other indicators, showing colour change from acidic to basic

medium)

(ci) no of moles of NaOH used to neutralize the acid

= 21.50/1000 x 0.203 = 0.00436 g

(cii) no of moles of H+(aq) reacted with Mg

= (0.5x25) / 1000 X2 0.00436 = 0.02064

No of moles of H2SO4 used = 0.2064/2 = 0.01032

(ciii) Mass of Mg reacted with the acid = 24.3x 0.01032 = 0.2508g

Percentage purity of Mg= 0.2508/0.27 X 100% = 92.89%

(di) The Mg ribbon may be coated with an oxide layer dur to exposure to air.

(dii) Clean the surface of the magnesium ribbon with sand paper beforeWeighing.

10.For the pH 2 HCl (aq), [H+ (aq)] = 0.01MNo of mole of HCl required for the preparation = 0.01X1.0 = 0.01

Mass of constant boiling HCl (aq) = (0.01X 36.5 )/ 0.202

= 1.80g

11.Mass of HNO3 in 1dm

3

= 1420 x 0.68- 965.6Concentration of the acid = 965.6/ 63 = 15.3M

12.(a) Mole ratio of C : H : O = 5.88: 5.93 :1.47= 4 : 4 : 1

Empirical formula of A = C4H4O

(b) Suppose the formula of this acid is HA

HA + NaOH NaA + H2O

1 mole 1 mole

No of moles of HA= no of moles of NaOH

0.3200/ M = 0.1 x 23.5/1000

M = 136

Thus the molecular mass of A is 136 g

68n = 136

n = 2

Molecular formula of A is C8H8O2

13.1014.C4H10O