CBSE Class 12 Chemistry Solutions Assignment Set B

Very Short Answer Type Questions

Question. What is meant by molality of a solution?
Answer: Molality of a solution can be defined as the number of moles of solute dissolved in one kg solvent. It is denoted by m.
m = Number of moles of solute / Mass of solvent in kg = n₂ / W₁

Question. State the main advantage of molality over molarity as the unit of concentration.
Answer: Molality is independent of temperature, whereas molarity is a function of temperature.

Question. Gas (A) is more soluble in water than gas (B) at the same temperature. Which one of the two gases will have the higher value of KH (Henry’s constant) and why?
Answer: According to Henry’s law, the solubility of a gas is inversely proportional to the Henry’s law constant (K_H) for that gas. Hence, gas (B) being less soluble, would have a higher K_H value.

Question. Explain the following : Henry’s law about dissolution of a gas in a liquid.
Answer: Henry’s law states that, the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.
p = K_H⋅x where, K_H = Henry’s law constant.
Different gases have different K_H values at the same temperature.

Question. Some liquids on mixing form ‘azeotropes’. What are ‘azeotropes’?
Answer: Azeotropes are the binary mixtures of solutions that have the same composition in liquid and vapour phases and that have constant boiling points. It is not possible to separate the components of azeotropes by fractional distillation.

Question. Defne the following term : Ideal solution
Answer: A solution which obeys Raoult’s law at all concentrations and temperatures is called an ideal solution.

Question. How is it that alcohol and water are miscible in all proportions
Answer: Both alcohol and water are polar in nature hence, they are miscible in all proportions. Water and ethanol molecules attract

each other because of the formation of H-bonds. This property also makes them miscible.

Question. State the condition resulting in reverse osmosis.
Answer: Reverse osmosis occurs when a pressure larger than the osmotic pressure is applied to the solution side.

Question. Defne the following term : Van’t Hoff factor
Answer: van’t How factor : It is defined as the ratio of the experimental value of colligative property to the calculated value of the colligative property and is used to find out the extent of dissociation or association. Mathematically, it is represented as

Short Answer Type Questions

Question. Calculate the temperature at which a solution containing 54 g of glucose, (C₆H₁₂O₆), in 250 g of water will freeze.
(K_f for water = 1.86 K mol^–1 kg)
Answer: M₂ (glucose, C₆H₁₂O₆) = 180 g mol^–1
W₂ = 54 g, W₁ = 250 g, K_f = 1.86 K kg mol^–1

Question. A solution containing 8 g of a substance in 100 g of diethyl ether boils at 36.86°C, whereas pure ether boils at 35.60°C. Determine the molecular mass of the solute. (For ether K_b = 2.02 K kg mol^–1)
Answer: T_b = 36.86°C, T°_b = 35.60°C
ΔT_b = T_b – T°_b = 36.86 – 35.60 = 1.26°C

Question. A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31°C. Determine the molar mass of this compound.
(B.pt. of pure benzene = 80.10°C and K_b for benzene = 2.53°C kg mol^–1)
Answer: W₂ = 1.25 g, W₁ = 99.0 g
ΔT_b = T_b – T°b = (80.31 – 80.10)°C = 0.21°C = 0.21 K
ΔT_b = K_b⋅m

Question. 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol^–1. Find the molar mass of the solute.
Answer: W₂ = 1.00 g, W₁ = 50 g, K_f = 5.12 K kg mol^–1,
ΔT_f = 0.40 K

Question. A 0.561 m solution of an unknown electrolyte depresses the freezing point of water by 2.93°C. What is van’t Hoff factor for this electrolyte?
The freezing point depression constant (K_f) for water is 1.86°C kg mol^–1.
Answer: m = 0.561 m, ΔT_f = 2.93°C and
K_f = 1.86°C kg mol^–1
ΔT_f = iK_f m

Question. Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2 K (the K_f for water = 1.86 K kg mo^l–1). 
Answer: ΔT_f = 2 K, K_f = 1.86 K kg mol^–1,
W₁ = 1 kg, ΔT_f = i K_f m, M₂(KCl) = 74.5 g mol^–1
i = 2 for KCl
ΔT_f = i x K_f x W₂ x 1000 / M₂ x W₁
2 = 2 x 1.86 x W₂ x 1000 / 74.5 x 1000
⇒ W₂ = 40.05 g

Question. (i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution?
Answer: (i) The elevation in boiling point of a solution is a colligative property which depends on the number of moles of solute added. Higher the concentration of solute added, higher will be the elevation in boiling point. Thus, 2 M glucose has higher boiling point than 1 M glucose solution.
(ii) When the external pressure applied becomes more than the osmotic pressure of solution then the solvent molecules from the solution pass through the semipermeable membrane to the solvent side and the process is called reverse osmosis.

Question. Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution?
Answer: (i) 1.2% sodium chloride solution is hypertonic with respect to 0.9% sodium chloride solution or blood cells thus, on placing blood cells in this solution exosmosis takes place that results in shrinking of cells.
(b) 0.4% sodium chloride solution is hypotonic with respect to 0.9% sodium chloride solution or blood cells thus, on placing blood cells in this solution endosmosis takes place that results in swelling of cells.

Question. Why does a solution containing non-volatile solute have higher boiling point than the pure solvent? Why is elevation of boiling point a colligative property?
Answer: The boiling point of the solution is always higher than that of the pure solvent. As the vapour pressure of the solution is lower than that of the pure solvent and vapour pressure increases with increase in temperature. Hence, the solution has to be heated more to make the vapour pressure equal to the atmospheric pressure. Elevation of boiling point is a colligative property because it depends on number of solute particles present in a solution.

Question. Calculate the mass of compound (molar mass = 256 g mol^–1) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K.
(Kf = 5.12 K kg mol^–1).
Answer: Given : W₂ = ?, M₂ = 256 g mol^–1, ΔT_f = 0.48 K
W₁ = 75 g, K_f = 5.12 K kg mol^–1

Question. 18 g of glucose, C₆H₁₂O₆ (Molar mass = 180 g mol^–1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil?
(K_b for water = 0.52 K kg mol^–1, boiling point of pure water = 373. 15 K)
Answer: Given W₁ = 1 kg = 1000 g, W₂ = 18 g,
M₂ = 180 g mol^–1
T°_b = 373.15 K, K_b = 0.52 K Kg mol^–1, T_b = ?
Using formula,

Question. Outer hard shells of two eggs are removed.
One of the egg is placed in pure water and the other is placed in saturated solution of sodium chloride. What will be observed and why?
Answer: The egg placed in pure water will swell because the concentration of proteins is high inside the egg as compared to water. Therefore, endosmosis occurs and water diffuses through the semipermeable membrane. The egg which is placed in sodium chloride solution will shrink due to osmosis of water out of the egg.

Question. Find the boiling point of a solution containing 0.520 g of glucose (C₆H₁2O₆) dissolved in 80.2 g of water.
[Given : K_b for water = 0.52 K/m]
Answer: Given, W₂ = 0.520 g,
W₁ = 80.2 g, Kb = 0.52 K m^–1
M₂ of C₆H₁₂O₆ = 6 × 12 + 12 × 1 + 6 × 16
= 180 g mol^–1

Question. What mass of ethylene glycol (molar mass = 62.0 g mol^–1) must be added to 5.50 kg of water to lower the freezing point of water from 0°C to – 10.0°C?
(Kf for water 1.86 K kg mol^–1)
Answer: M₂ (ethylene glycol) = 62 g mol^–1
W₁ = 5.50 kg = 5500 g, ΔT_f = T°_f – T_f = 0°C – (–10°C)
= 10°C = 10 K and K_f = 1.86 K kg mol^–1

Question. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Answer: Given : p°_A = 450 mm Hg, p°_B = 700 mm Hg,
PTotal = 600 mm Hg, x_A = ?
Applying Raoult’s law, p_A = x_A × p°_A
p_B = x_B × p°_B = (1 – xA)p°_B
P_Total = p_A + p_B = x_A × p°A + (1 – x_A)p°_B
= p°_B + (p°_A – p°_B)x_A
Substituting the given values, we get
600 = 700 + (450 – 700)x_A or, 250x_A = 100
or x_A = 100/250 = 0.40

Question. The partial pressure of ethane over a saturated solution containing 6.56 × 10^–2 g of ethane is 1 bar. If the solution contains 5.0 × 10^–2 g of ethane, then what will be the partial pressure of the gas?
Answer: Applying the relationship,
m = K_H × p
In the first case, 6.56 × 10^–2 g = K_H × 1 bar
or, K_H = 6.56 × 10^–2 g bar^–1
In the second case,
5.0 × 10^–2 g = (6.56 × 10^–2 g bar^–1) × p

P = 5.0 × 10^–2 g / 6.56 × 10^–2 g bar^–1 = 0.762 bar

 Question. Calculate the freezing point of solution when 1.9 g of MgCl₂(M = 95 g mol^–1) was dissolved in 50 g of water, assuming MgCl₂ undergoes complete ionization.
(Kf for water = 1.86 K kg mol^–1)
Answer: 

Long Answer Type Questions

 Question. An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₄(OH)₂) and 200 g of water. Calculate the molality of the solution.
If the density of this solution be 1.072 g mL^–1 what will be the molality of the solution?
Answer: Mass of the solute, C₂H₄(OH)₂ = 222.6 g
Molar mass of solute, C₂H₄(OH)₂ = 62 g mol^–1

Question. When 2.56 g of sulphur was dissolved in 100 g of CS₂, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx).
(K_f the CS₂ = 3.83 K kg mol^–1, atomic mass of sulphur = 32 g mol^–1)
Answer: W₂ = 2.56 g, W₁ = 100 g, ΔT_f = 0.383 K
K_f = 3.83 K kg mol^–1, ΔT_f = K_f × m

Question. Calculate the boiling point of solution when 4 g of MgSO₄ (M = 120 g mol^–1) was dissolved in 100 g of water, assuming MgSO₄ undergoes complete ionization.
(K_b for water = 0.52 K kg mol^–1)
Answer: W₂ = 4 g, M₂ = 120 g mol^–1
W₁ = 100 g, K_b = 0.52 K kg mol^–1
For complete dissociation, i = 2
Using formula, ΔTb = iK_bm

Question. 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 mol^–1)
(K_f for water = 1.86 K kg mol^–1)
Answer: W₂ = 10.50 g, W₁ = 200 g
M₂(MgBr₂) = 184 g mol^–1
K_f = 1.86 K kg mol^–1

Question. Phenol associates in benzene to a 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 K_f for benzene = 5.1 K m^–1]
Answer: Here, n = 2 because phenol forms dimer on association.
W₂ = 20 g, W₁ = 1 kg = 1000 g, ΔT_f = 0.69 K,
K_f = 5.1 K m^–1

Question. The boiling point elevation of 0.30 g acetic acid in 100 g benzene is 0.0633 K. Calculate the molar mass of acetic acid from this data. What conclusion can you draw about the molecular state of the solute in the solution?
[Given K_b for benzene = 2.53 K kg mol^–1]
Answer: 

Question. The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g of benzene is lowered by 0.45°C. Calculate.
(i) the molar mass of acetic acid from this data
(ii) van’t Hoff factor
[For benzene, K_f = 5.12 K kg mol^–1] What conclusion can you draw from the value of van’t Hoff factor obtained?
Answer: 

Question. Calculate the freezing point of the solution when 31 g of ethylene glycol (C₂H₆O₂) is dissolved in 500 g of water.
(K_f for water = 1.86 K kg mol^–1)
Answer: Mass of ethylene glycol (C₂H₆O₂), W₂ = 31 g
Mass of water, W1 = 500 g
M₂ (Mol. mass of C₂H₆O₂) = 62 g mol^–1,
K_f = 1.86 K kg mol^–1, T_f = ?

Question. A 5% solution (by mass) of cane-sugar in water has freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water if the freezing point of pure water is 273.15 K.
[Molecular masses : Glucose C₆H₁₂O₆ : 180 amu;
Cane-sugar C₁₂H₂₂O₁₁ : 342 amu]
Answer: Molality of sugar solution

Question. 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 while pure water boils at 100°C. What mass of glycerol was dissolved to make the solution?
(K_b for water = 0.512 K kg mol^–1)
Answer: W₁ = 500 g
Boiling point of solution (T_b) = 100.42°C
K_b for water = 0.512 K kg mol^–1
M₂(C₃H₈O₃) = (3 × 12) + (8 × 1) + (3 × 16)
= 92 g mol^–1
ΔTb = Tb – T° _b
= 373.42 K – 373 K = 0.42 K

Question. 15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at –0.34°C. What is the molar mass of this material?
(K_f for water = 1.86 K kg mol^–1)
Answer: W₁ = 450 g, W₂ = 15.0 g
ΔT_f = T°_f – T_f = 273 K – 272.66 K = 0.34 K
K_f = 1.86 K kg mol^–1

Question. A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is added to this solution. The new vapour pressure becomes 2.9 kPa at 298 K. Calculate (i) the molecular mass of solute and (ii) vapour pressure of water at 298 K.
Answer: The relative lowering of vapour pressure is given by the following expression,
(p°solvent – psolution)/p°solvent = n₂/(n₁ + n₂)
for dilute solutions, n₂ << n₁, therefore
(p°solvent – psolution)/p°solvent = n₂/n₁
= (W₂ × M₁)/(M₂ × W₁)
(p°solvent – 2.8)/p°solvent = (30 × 18)/(M₂ × 90)
(p°solvent – 2.8)/p°solvent = 6/M₂ …(1)
Similarly for second case we get,
(p°solvent – 2.9)/p°solvent = (30 × 18)/(M₂ × 108)
(p°solvent – 2.9)/p°solvent = 5/M₂ …(2)
On solving eq. (1) and (2), we get
(p°solvent – 2.8)/(p°solvent – 2.9) = 6/5
∴ p°solvent = 3.4 kPa
i.e., vapour pressure of water at 298 K is 3.4 kpa Substituting the value of p°solvent in (1) we get,
(3.4 – 2.8)/3.4 = 6/M₂
or 0.6/3.4 = 6/M₂
∴ M₂ = 34 g

Question. Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.00 g of water. (K_b for water = 0.512 K kg mol^–1),
(Molar mass of NaCl = 58.44 g)
Answer: i = 2, K_b = 0.512 K kg mol^–1, WB = 15 g
M_B = 58.44 g mol^–1, W_A = 250 g

Therefore, boiling point of aqueous solution,
T_b = T°_b + ΔT_b = 373.15 K + 1.05 K = 374.20 K

Question. Calculate the boiling point of one molar aqueous solution (density 1.06 g ml^–1) of KBr.
[Given : K_b for H₂O = 0.52 K kg mol^–1, atomic  
mass : K = 39, Br = 80]
Answer: Concentration of the solution = 1 molar
Density of the solution = 1.06 g mL^–1
M₂, molar mass of KBr = 39 + 80 = 119 g mol^–1
K_b for H₂O = 0.52 K kg mol^–1

Question. Calculate the freezing point of a solution containing 18 g glucose, C₆H₁₂O₆ and 68.4 g sucrose, C₁₂H₂₂O₁₁ in 200 g of water. The freezing point of pure water is 273 K and Kf for water is 1.86 K m^–1.
Answer: Molar mass of glucose, C₆H₁₂O₆
= 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol^–1
Molar mass of sucrose, C₁₂H₂₂O₁₁
= 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol^–1

Question. A 0.1539 molal aqueous solution of cane sugar (mol. mass = 342 g mol^–1) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of glucose (mol. mass = 180 g mol^–1) per 100 g of solution.
Answer: Molality = 0.1539 m,
ΔT_f = T°_f – T_f = 273.15 – 271 = 2.15 K

Question. If N₂ gas is bubbled through water at 293 K, how many millimoles of N₂ gas would dissolve in 1 litre of water? Assume that N₂ exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N₂ at 293 K is 76.48 k bar.
Answer: According to Henry’s law, P_N2 = K_H × X_N2

Thus, composition of the liquid mixture will be
x_A = 0.40
x_B = 1 – 0.40 = 0.60

Calculation of composition in the vapour phase,
p_A = x_A × p°_A = 0.40 × 450 mm Hg = 180 mm Hg
p_B = x_B × p°_B = 0.60 × 700 mm Hg = 420 mm Hg
Mole fraction of A in the vapour phase
= P_A/P_A+P_B = 180/180+420 = 0.30
Mole fraction of B in the vapour phase = -  0.30
= 0.70