WHICH IS MORE WHICH IS MORE CONCENTRATED ???CONCENTRATED ???

BEAKER ABEAKER A BEAKER BBEAKER B

CONCENTRATIONCONCENTRATIONOFOF

ACIDS & ALKALISACIDS & ALKALIS

Quantity of solute in a given Quantity of solute in a given volume of solution, which is volume of solution, which is usually 1 dmusually 1 dm33

CONCENTRATIONCONCENTRATION

REMEMBER THIS FORMULAE!!!REMEMBER THIS FORMULAE!!!

Concentration = mass of solute (g)Concentration = mass of solute (g)

volume of solution volume of solution (dm(dm33))

Unit : g dmUnit : g dm-3-3

1 dm1 dm33 = 1000 cm = 1000 cm33

MOLARITYMOLARITY

is the number of moles of solute that are present in 1 dm-3 of solution

Molarity = number of moles of solute Molarity = number of moles of solute (mol)(mol)

volume of solution (dmvolume of solution (dm33))

Unit: mol dmUnit: mol dm-3 -3

REMEMBER THIS FORMULAE!!!REMEMBER THIS FORMULAE!!!

Molarity ConcentrationMolarity Concentration

(mol dm(mol dm-3-3) (g dm) (g dm-3-3))

REMEMBER THIS FORMULAE...!!!REMEMBER THIS FORMULAE...!!!

X Molar mass

÷ Molar mass

Molarity = number of moles of solute Molarity = number of moles of solute (mol)(mol)

volume of solution (dmvolume of solution (dm33))

## no. of mole (n) = molarity of solution no. of mole (n) = molarity of solution (M) (M) X volume (V) X volume (V)

REMEMBER THIS FORMULAE!!!REMEMBER THIS FORMULAE!!!

In short….

n = no. of moles of solute (mol)

M = molarity of solution (mol dm-3)

V = volume of solution (dm3)

n = MV

No of moles = mass of solute (g)No of moles = mass of solute (g)

molar mass (g molmolar mass (g mol--11))

Find the concentration of a solution in grams per dm3 when 36.5 g of hydrogen chloride, HCl is dissolved in water to make up 500 cm3 of solution.

QUESTION 1QUESTION 1

QUESTION 2QUESTION 2

5.00 g of copper (II) sulphate is dissolved in water to form 500 cm3 solution. Calculate the concentration of copper (II) sulphate solution in g dm-3

What is the mass of sodium carbonate required to dissolve in water to prepare a 200 cm3 solution that contains 50 g dm-3 ?

QUESTION 3QUESTION 3

QUESTION 4QUESTION 4

A student dissolves 50.0 g of copper (II) sulphate in water to make a 250 cm3 of solution. What is the concentration of the solution in

g dm-3 ?

Find the molarity of a solution which is prepared by dissolving 0.30 mol of sodium hydroxide, NaOH in distilled water to make up 250 cm3 of solution.

QUESTION 5QUESTION 5

28.0 g of potassium hydroxide is dissolved in water to make a 200 cm3 of solution. Calculate the molarity of potassium hydroxide solution obtained.

[ RAM: H, 1; O, 16; K, 39]

QUESTION 6QUESTION 6

0.1 mole of zinc nitrate is dissolved in water to produce a solution with a molarity of 0.2 mol dm-3. What is the volume of zinc nitrate solution?

QUESTION 7QUESTION 7

The molarity of a bottle of nitric acid,HNO3 solution is 2.0 mol dm-3. What is the concentration of the solution in g dm-3 ?

[Relative atomic mass: H, 1; N, 14; O, 16]

QUESTION 8QUESTION 8

Calculate the molarity of a sodium sulphate, Na2SO4 solution with a concentration of 28.4 g dm-3 .

[ Relative atomic mass: O, 16; Na, 23; S, 32 ]

QUESTION 9QUESTION 9

A potassium chloride solution has a concentration of 14.9 g dm-3. What is the molarity of this solution in mol dm-

3 ?

[ Relative atomic mass: Cl, 35.5; K, 39 ]

QUESTION 10QUESTION 10

The molarity of barium hydroxide solution in a reagent bottle is 0.4 mol dm-3. What is the concentration of this solution in g dm-3?

[ Relative atomic mass: H, 1; O, 16; Ba, 137 ]

QUESTION 11QUESTION 11

4.0 g of sodium carbonate powder, Na2CO3 is dissolved in water and made up to 250 cm3. What is the molarity of the sodium carbonate solution?

[ Relative atomic mass: C, 12; O, 16; Na, 23 ]

QUESTION 12QUESTION 12

Calculate the mass of calcium hydroxide in 50 cm3 of 0.01 mol dm-3 calcium hydroxide solution.

[ Relative atomic mass: H, 1; O, 16; Ca, 40 ]

QUESTION 13QUESTION 13

A 0.5 mol dm-3 lithium hydroxide contains 6.0 g lithium hydroxide. Calculate the volume of this solution.

[ Relative atomic mass: H, 1; Li, 7; O, 16 ]

QUESTION 14QUESTION 14

PREPARATION PREPARATION OF STANDARD OF STANDARD

SOLUTIONSSOLUTIONS

Standard solution is a solution that we had already knew its concentration.

Volumetric flask is a apparatus with certain volume to use preparation standard solution

STANDARD SOLUTION

STEP IN THE PREPARATION STEP IN THE PREPARATION OF A STANDARD SOLUTIONOF A STANDARD SOLUTION

STEP 1STEP 1- Calculate the mass of solute needed.Calculate the mass of solute needed.

STEP 2STEP 2- Weigh out the exact mass of solute Weigh out the exact mass of solute

needed.needed.

STEP 3STEP 3

- Dissolve the solute in a small amount - Dissolve the solute in a small amount of distilled water.of distilled water.

STEP 4STEP 4- Transfer the dissolved solute

into a suitable volumetric flask.

STEP 5STEP 5

- Add enough water to the required volume.

PREPARATION PREPARATION OF A SOLUTION OF A SOLUTION

BY DILUTION BY DILUTION METHODMETHOD

DI LUTI ON

preparation

f ormula

meaning

MEANING OF MEANING OF DILUTIONDILUTION

is a process of diluting a concentrated solution by adding a solvent such as water to obtain a more diluted solution.

STEP IN THE PREPARATION STEP IN THE PREPARATION OF A SOLUTION BY DILUTION OF A SOLUTION BY DILUTION METHODMETHOD

Calculate the volume of stock Calculate the volume of stock solution required.solution required.

Pipette the required volume of Pipette the required volume of stock solution into a volumetric stock solution into a volumetric flask.flask.

Add water to the required Add water to the required volumevolume

REMEMBER THIS REMEMBER THIS FORMULAE…!!!FORMULAE…!!!

M1 V1 = M2 V2

1000 1000

M1 V1 = M2 V2

M1 = concentration BEFORE dilutionV1 = volume BEFORE dilutionM2 = concentration AFTER dilutionV2 = volume AFTER dilution

CALCULATI ONS ON MOLARI TY

M1 = Initial molarityM2 = Final molarityV1 = Initial volumeV2 = Final volume

Thus the formula

M1V1 = M2V2can be used to find the

new molarity

ANSWER THIS…ANSWER THIS…

what is the volume of 2.0 mol dm-3 nitric acid needed to be diluted with distilled water to make 250 cm3 of 0.5 mol dm-

3 nitric acid ?

SOLUTION….SOLUTION….Before dilution : M1 = 2.0 mol dm-3

V1 = ?

After dilution : M2 = 0.5 mol dm-3

V2 = 250 cm3

Using formula M1V1 = M2V2

(2.0 mol dm-3) V1 = 0.5 mol dm-3 (250 cm3)

V1 = 0.5 mol dm-3 (250 cm3) 2.0 mol dm-3

V1 = 62.5 cm3

QUESTION 2QUESTION 2

50 cm3 of water is added to 200 cm3 of a 2 mol dm-3 solution of NaOH. Determine the molarity of the diluted solution.

QUESTION 3QUESTION 3

What is the volume of 1.0 mol dm-3 HCl is required to be diluted with distilled water to produce 100 cm3 of 0.1 mol dm-3 solution of HCl?