UNITY PUBLIC SCHOOL

SOLUTION - WORK SHEET 1

CHEMISTRY
Class : XII

Roll No. : Time -

Date : 18/07/2023 MM - 100

1. A 5% solution by mass of cane sugar, C12H22O11 (molecular weight 342) is isotonic with 0.877% 3
solution of substance ‘X’. Find the molecular weight of substance X.

Ans :

2. 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate density and molality of 3
KOH solution. [K = 39, O = 16, H = 1]

Ans :

3. What is the molality of ammonia in a solution containing 0.85 g of NH3 in 100 mL of a liquid of 3
density 0.85 g cm–3?

Ans :

4. What is the mass of precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with 50 3
mL of 5.8% solution of NaCl?
[Ag = 108.0, N = 14, O = 16, Na = 23, Cl = 35.5]

Ans :

5. The Henry’s Law constant for oxygen dissolved in water is 4.34×104 atm at 25 °C. If the partial 3
pressure of oxygen in air is 0.2 atm, calculate the solubility of oxygen in water at 25 °C.
Ans :

6. Vapour pressure of water at 20 °C is 17.5 mm Hg, Calculate the vapour pressure of water at 20 3
°C when 15 g glucose (molecular weight 180 g mol–1) is dissolved in 150 g of water.

Ans :

7. A solution is prepared by dissolving 5 g non-voltale solute in 95 g of water. It has vapour 3

pressure of 23.375 mm of Hg at 25 °C. Calculate the molar mass of solute. (Vapour pressure of
pure water at 25 °C = 23.75 mm of Hg)

Ans :

8. Calculate the normal boiling point of a sample of sea water containing 3.5% of NaCl and 0.13% 3
of MgCl2 by mass.

[Given Kb (water) = 0.52 K kg mol–1, Mol. Wt. of NaCl = 58.5 g mol–1, MgCl2 = 95 g mol–1].
Ans : Assuming complete dissociation of NaCl and MgCl2, i.e. 1 mole of NaCl produces 2
moles of species and 1 mole of MgCl2 produces 3 moles of species.

9. An aqueous solution of 3.12 g of BaCl2 in 250 g of water is found to boil at 100.0832 °C. 3
Calculate the degree of dissociation of BaCl2. [Kb (H2O) = 0.52 K/m.]

Ans :

10. Calculate the freezing point of a 1 molar aqueous solution of KCl. 3

(Density of solution = 1.04 g cm–3, Kf = 1.86 K kg mol–1, At. Wt. of K = 39 and Cl = 35.5)

Ans :

11. (a) Calculate the molality of sulphuric acid solution in which mole fraction of water is 0.8. [H = 1, 5
S = 32, O = 164]
(b) Calculate molality and mole fraction of solute in a sugar syrup of mass 214.2 g containing 34.2
g of sucrose (MWt of sucrose, C12H22O11 = 342 g mol–1).
Ans :

12. (a) A sample of water was found to contain dissolved oxygen (O2) to the extent of 5 ppm and 5
hardness due to Mg2+ is 15 ppm. Calculate the amount of O2 and number of Mg2+ in 1 litre of water
(density of water = 1g/mL).

(b) What volume of 98% sulphuric acid (d = 1.84 g cm–3) and what mass of water must be required
to prepare 500 cm3 of 15% solution of H2SO4 (d = 1.10 g cm–3)? [H = 1, S = 32, O = 16 u]
Ans :

13. (a) State Raoult’s law for a solution containing volatile components. Name the solution which 5
follows Raoult’s law at all concentrations and temperatures.

(b) Calculate the boiling point elevation for a solution prepared by adding 10 g of CaCl2 to 200 g of
water. (Kb for water = 0.512 K kg mol–1, Molar mass of CaCl2 = 111 g mol–1)

Ans : (a) Raoult’s law: It states that vapour pressure of each component is directly proportional
to its mole fraction when both solute and solvent are volatile.

Ideal solution follows Raoult’s law at all concentrations and temperatures.

(b)

14. (a) Define the following terms: 5

(i) Azeotrope (ii) Osmotic pressure (iii) Colligative properties
(b) Calculate the molarity of 9.8% (w/w) solution of H2SO4 if the density of the solution is 1.02 g
mL–1 (Molar mass of H2SO4 = 98 g mol–1)
Ans : (a) (i) Azeotrope: Those solutions which distill out unchanged in their composition are
called azeotropes. They are constant boiling mixtures.
(ii) Osmotic pressure: It is an extra pressure which must be applied on solution side so
as to stop the flow of solvent molecules into solution when both are separated by a
semipermeable membrane.
(iii) Colligative property: The property which depends on the number of particles of
solute and not on the nature of solute is called colligative property.

(b)

15. (a) Define the following terms: 5

(i) Molarity
(ii) Molal elevation constant (Kb)
(b) A solution containing 15 g urea (molar mass = 60 g mol–1) per litre of solution in water has the
same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol–1) in water.
Calculate the mass of glucose present in one litre of its solution.

Ans : (a) (i) Molarity: It is defined as the number of moles of solute dissolved per litre of
solution.
(ii) Molal elevation constant (Kb): It is equal to elevation in boiling point when solution is
one molal.

(b)

16. (a) What is van’t Hoff factor? What possible values can it have if the solute molecules undergo 5
dissociation?
(b) An aqueous solution containing 12.48 g of barium chloride in 1.0 kg of water boils at 373.0832
K. Calculate the degree of dissociation of barium chloride.
[Given; Kb for H2O = 0.52 K m–1; Molar mass of BaCl2 = 208.34 g mol–1]
Ans :

17. (a) Explain the following: 5

(i) Henry’s law about dissolution of a gas in a liquid.
(ii) Boiling point elevation constant for a solvent.
(b) A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of
water. This solution has a boiling point of 100.42 °C. What mass of glycerol was dissolved to make
this solution? (Kb for water = 0.512 K kg mol–1)

Ans : (a) (i) Henry’s Law: It states that the solubility of a gas in a liquid is directly proportional
to the pressure of the gas. If mole fraction of a gas in the solution is used as a measure
of solubility, then it can be defined as the mole fraction of a gas in the solution is directly
proportional to the partial pressure of the gas over the solution.
p = KH x,
where ‘p’ is partial pressure of gas, ‘x’ is mole fraction of the gas and KH is Henry’s law
constant.
(ii) Boiling Point Elevation Constant (Molal Boiling Point Elevation Constant): It is equal to
elevation in boiling point of 1 molal solution, i.e. 1 mole of solute is dissolved in 1 kg of
solvent. The units of Kb is K/m or °C/m or K kg mol–1.

(b)

18. (a) Differentiate between molarity and molality for a solution. How does a change in temperature 5
influence their values?
(b) Calculate the freezing point of an aqueous solution containing 10.50 g of MgBr2 in 200 g of
water. (Molar mass of MgBr2 = 184 g) (Kf for water = 1.86 K kg mol–1)
Ans : (a) Molality is defined as the number of moles of solute per kg of solvent. It is not
affected by temperature. Molarity is defined as the number of moles of solute per litre of
solution. It decreases with increase in temperature.

(b)

19. (a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a 5
colligative property? Explain.
(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water.
(Kb for water = 0.512 K kg mol–1, Molar mass of NaCl = 58.44 g)

Ans : (a) Osmosis: When a solution is separated from the solvent by a semipermeable
membrane which allows the passage of solvent molecules but does not allow solute
particles to pass through it, there is net flow of solvent molecules from the solvent to the
solution which is called osmosis.
Osmotic Pressure: Osmotic pressure may be defined as an extra pressure that must be
applied to the solution side to prevent the flow of solvent into solution through a
semipermeable membrane.

(b)

20. (a) State the following: 5

(i) Henry’s law about partial pressure of a gas in a mixture.
(ii) Raoult’s law in its general form in reference to solutions.
(b) A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an
osmotic pressure of 0.335 torr at 25 °C. Assuming the gene fragment is non-electrolyte, determine
its molar mass.
Ans : (a) (i) Henry’s Law: It states that the partial vapour pressure of a gas in vapour phase is
directly proportional to mole fraction of the gas in the solution.

where KH is Henry’s law constant, ‘x’ is mole fraction of the gas in solution and p is
partial vapour pressure of the gas in solution.

(ii) Raoult’s law for solution of non-volatile solute: The relative lowering of vapour
pressure for a solution is equal to the mole fraction of solute when solvent alone is
volatile.

21. (a) List any four factors on which the colligative properties of a solution depend. 5
(b) Calculate the boiling point of one molar aqueous solution (density 1.06 gmL–1) of KBr.
[Given: Kb for H2O = 0.52 K kg mol–1, Atomic mass: K = 39, Br = 80]

Ans : (a) (i) Number of particles of solute

(ii) Association or dissociation of solute
(iii) Concentration of solution
(iv) Temperature

(b)
22. (a) Explain why a solution of chloroform and acetone shows negative deviation from Raoult’s 5
law.
(b) Phenol associates in benzene to certain extent to form a dimer. A solution containing 20 g of
phenol in 1.0 kg of benzene has its freezing point lowered by 0.69 K. Calculate the fraction of
phenol that has dimerised. [Given Kf for benzene = 5.1 Km–1]

Ans : (a) It is due to the formation of H-bonding due to which escaping tendency of molecules
and vapour pressure of the solution decrease, and boiling point of the solution increases.
Hence, such solution shows negative deviation from Raoult’s law.

(b)

23. (a) Define the terms osmosis and osmotic pressure. What is the advantage of using osmotic 5
pressure as compared to other colligative properties for the determination of molar masses of
solutes in solutions?
(b) A solution prepared from 1.25 g of oil of wintergreen (methyl salicylate) in 90.0 g of benzene
has a boiling point of 80.31 °C. Determine the molar mass of this compound. (Boiling point of pure
benzene = 80.10 °C and Kb for benzene = 2.53 °C kg mol–1)

Ans : (a) Osmosis: When a solution is separated from the solvent by a semipermeable
membrane which allows the passage of solvent molecules but does not allow solute
particles to pass through it, there is net flow of solvent molecules from the solvent to the
solution which is called osmosis.
Osmotic Pressure: Osmotic pressure may be defined as an extra pressure that must be
applied to the solution side to prevent the flow of solvent into solution through a
semipermeable membrane. Osmotic pressure is determined at room temperature and
has appreciable value which can be easily measured.

(b)

24. (a) Assuming complete ionisation, calculate the expected freezing point of solution prepared by 5
dissolving 6.00 g of Glauber’s salt, Na2SO4.10H2O in 0.1 kg of H2O. (Kf for H2O = 1.86 K kg mol–1)
[At. mass of Na = 23, S = 32, O = 16, H = 1 u].
(b) Two liquids X and Y boil at 110 °C and 130 °C respectively. Which of them has higher vapour
pressure at 50 °C ?
Ans : (a)

(b) X has lower boiling point, therefore, it will vaporise easily and will have higher vapour
pressure than Y at 50 °C.