Chapter 2 Solutions

Maharashtra State Board 12th Chemistry Solutions Chapter 2 Solutions

1. Choose the most correct answer.

Question i.
The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is
(a) 24 mm Hg
(b) 32 mm Hg
(c) 48 mm Hg
(d) 12 mm Hg
Answer:
(d) 12 mm Hg

Question ii.
The colligative property of a solution is
(a) vapour pressure
(b) boiling point
(c) osmotic pressure
(d) freezing point
Answer:
(c) osmotic pressure

Question iii.
In calculating osmotic pressure the concentration of solute is expressed in
(a) molarity
(b) molality
(c) mole fraction
(d) mass per cent
Answer:
(a) molarity

Question iv.
Ebullioscopic constant is the boiling point elevation when the concentration of solution is
(a) 1 m
(b) 1 M
(c) 1 mass%
(d) 1 mole fraction of solute
Answer:
(a) 1 m

Question v.
Cryoscopic constant depends on
(a) nature of solvent
(b) nature of solute
(c) nature of solution
(d) number of solvent molecules
Answer:
(a) nature of solvent

Question vi.
Identify the correct statement
(a) vapour pressure of solution is higher than that of pure solvent.
(b) boiling point of solvent is lower than that of solution
(c) osmotic pressure of solution is lower than that of solvent
(d) osmosis is a colligative property.
Answer:
(b) boiling point of solvent is lower than that of solution

Question vii.
A living cell contains a solution which is isotonic with 0.3 M sugar solution. What osmotic pressure develops when the cell is placed in 0.1 M KCl solution at body temperature ?
(a) 5.08 atm
(b) 2.54 atm
(c) 4.92 atm
(d) 2.46 atm
Answer:
(c) 4.92 atm

Question viii.
The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose isotonic with blood has the percentage (by volume)
(a) 5.41%
(b) 3.54%
(c) 4.53%
(d) 53.4%
Answer:
(a) 5.41%

Question ix.
Vapour pressure of a solution is
(a) directly proportional to the mole fraction of the solute
(b) inversely proportional to the mole fraction of the solute
(c) inversely proportional to the mole fraction of the solvent
(d) directly proportional to the mole fraction of the solvent
Answer:
(d) directly proportional to the mole fraction of the solvent

Question x.
Pressure cooker reduces cooking time for food because
(a) boiling point of water involved in cooking is increased
(b) heat is more evenly distributed in the cooking space
(c) the higher pressure inside the cooker crushes the food material
(d) cooking involves chemical changes helped by a rise in temperature
Answer:
(a) boiling point of water involved in cooking is increased

Question xi.
Henry’s law constant for a gas CH₃Br is 0.159 mol dm^-3atm at 250°C. What is the solubility of CH₃Br in water at 25 °C and a partial pressure of 0.164 atm?
(a) 0.0159 mol L^-1
(b) 0.164 mol L^-1
(c) 0.026 M
(d) 0.042 M
Answer:
(c) 0.026 M

Question xii.
Which of the following statement is NOT correct for 0.1 M urea solution and 0.05 M sucrose solution ?
(a) osmotic pressure exhibited by urea solution is higher than that exhibited by sucrose solution
(b) urea solution is hypertonic to sucrose solution
(c) they are isotonic solutions
(d) sucrose solution is hypotonic to urea solution
Answer:
(c) they are isotonic solutions

2. Answer the following in one or two sentences

Question i.
What is osmotic pressure ?
Answer:
(1) Definition : The osmotic pressure is defined as the excess mechanical pressure required to be applied to a solution separated by a semipermeable membrane from pure solvent or a dilute solution to prevent the osmosis or free passage of the solvent molecules at a given temperature. The osmotic pressure is a colligative property.

Osmosis and osmotic pressure

(2) Explanation : Consider an inverted thistle funnel on the mouth of which a semipermeable membrane is firmly fastened. It is filled with the experimental solution and immersed in a solvent like water. As a result, solvent molecules pass through the membrane into the solution in the funnel causing rising of level in the arm of thistle funnel. This increases the hydrostatic pressure. At a certain stage this rising level stops indicating an equilibrium between the rates of flow of solvent molecules from solvent to solution and from solution to solvent. The hydrostatic pressure at this stage represents osmotic pressure of the solution in the thistle funnel.

Question ii.
A solution concentration is expressed in molarity and not in molality while considering osmotic pressure. Why ?
Answer:

Question iii.
Write the equation relating boiling point elevation to the concentration of solution.
Answer:
The elevation in the boiling point of a solution is directly proportional to the molal concentration (expressed in mol kg^-1) of the solution.
Hence, if ΔT_bis the elevation in the boiling point of a solution of molal concentration m then,
ΔT_b∝ m
∴ ΔT_b= K_bm
where K_bis a proportionality constant.
If m = 1 molal,
ΔT_b= K_b
K_bis called the ebullioscopic constant or molal elevation constant. K_bis characteristic of the solvent.

Question iv.
A 0.1 m solution of K₂SO₄in water has freezing point of -0.43 °C. What is the value of van’t Hoff factor if K_ffor water is 1.86 K kg mol^-1?
Answer:
Given : m = 0.1 m, ΔT_f= 0 – (-0.43) = 0.43 °C
K_f= 1.86 K kg mol^-1, i = ?
ΔT_f= i × K_f× m
∴ i = = 2.312
van’t Hoff factor = i = 2.312

Question v.
What is van’t Hoff factor?
Answer:
Definition of the van’t Hoff factor, i : It is defined as a ratio of the observed colligative property of the solution to the theoretically calculated colligative property of the solution without considering molecular change.

The van’t Hoff factor can be represented as,

This colligative property may be the lowering of vapour pressure of a solution, the osmotic pressure, the elevation in the boiling point or the depression in the freezing point of the solution. Hence,

From the value of the van’t Hoff factor, the degree of dissociation of electrolytes, degree of association of nonelectrolytes can be obtained.

van’t Hoff factor gives the important information about the solute molecules in the solution and chemical bonding in them.

Question vi.
How is van’t Hoff factor related to degree of ionization?
Answer:
Consider 1 dm³of a solution containing m moles of an electrolyte AxBy. The electrolyte on dissociation gives x number of A^y+ions and y number of B^x-ions. Let α be the degree of dissociation.

At equilibrium,
AxBy ⇌ xA^y++ yB^x-
For 1 mole of electrolyte : 1 – α,  xα,  yα
and For ‘m’ moles of an electrolyte : m(1 – α), mxα, myα are the number of particles.
Total number of moles at equilibrium, will be,
Total moles = m(1 – α) + mxα + myα
= m[(1 – α) + xα + yα]
= m[1 + xα + yα – α]
= m[1 + α(x + y – 1)]

The van’t Hoff factor i will be,

If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n
∴ i = 1 + α(n – 1)
∴ α(n – 1) = i – 1
∴ α = ……(1)
This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.

Question vii.
Which of the following solutions will have higher freezing point depression and why ?
a. 0.1 m NaCl b. 0.05 m Al₂(SO₄)₃
Answer:
(1) Freezing point depression is a colligative property, hence depends on the number of particles in the solution.
(2) More the number of particles in the solution, higher is the depression in freezing point.
(3) The number of particles (ions) from electrolytes are,

(4) Therefore Al₂(SO₄)₃solution will have higher freezing point depression.

Question viii.
State Raoult’s law for a solution containing a nonvolatile solute.
Answer:
Statement of Raoult’s law : The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by mole fraction of the solvent at constant temperature.

Question ix.
What is the effect on the boiling point of water if 1 mole of methyl alcohol is added to 1 dm³of water? Why?
Answer:

Question x.
Which of the four colligative properties is most often used for molecular mass determination? Why?
Answer:

3. Answer the following.

Question i.
How vapour pressure lowering is related to a rise in boiling point of solution?
Answer:
(1) The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure, generally 1 atm (101.3 × 10³Nm^-2).

(2) When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure.
If the liquid has a low vapour pressure, it has a higher boiling point.

Vapour pressure curve showing elevation in boiling point

(3) When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases.
This is explained by graphical representation of the vapour pressure and the boiling point of the pure solvent and the solution.

If T₀and T are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by,
ΔT_b= T – T₀
The curve AB, represents the variation in the vapour pressure of a pure solvent with temperature while curve CD represents the variation in the vapour pressure of the solution.

(4) This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., P₀– P, where P₀and P are the vapour pressures of the pure solvent and the solution.
[ΔT_b∝ (P₀– P) or ΔT_b∝ ΔP]

Question ii.
What are isotonic and hypertonic solutions?
Answer:
(1) Isotonic solutions : The solutions having the same osmotic pressure at a given temperature are called isotonic solutions.

Explanation : If two solutions of substances A and B contain n_Aand n_Bmoles dissolved in volume V (in dm³) of the solutions, then their concentrations are,
C_A= (in mol dm^-3) and
C_B= (in mol dm^-3)

If the absolute temperature of both the solutions is T, then by the van’t Hoff equation,
π_A= C_ART and π_B= C_BRT, where π_Aand π_Bare their osmotic pressures.
For the isotonic solutions,
π_A= π_B
∴ C_A= C_B
∴
∴ n_A= n_B
Hence, equal volumes of the isotonic solutions at the same temperature will contain equal number of moles (hence, equal number of molecules) of the substances.

(2) Hypertonic solutions : When two solutions have different osmotic pressures, then the solution having higher osmotic pressure is said to be a hypertonic solution with respect to the other solution.

Explanation : Consider two solutions of substances A and B having osmotic pressures π_Aand π_B. If π_Bis greater than π_A, then the solution B is a hypertonic solution with respect to the solution A.
Hence, if C_Aand C_Bare their concentrations, then C_B> C_A. Hence, for equal volume of the solutions, n_B> n_A.

Question iii.
A solvent and its solution containing a nonvolatile solute are separated by a semipermable membrane. Does the flow of solvent occur in both directions? Comment giving reason.
Answer:

Question iv.
The osmotic pressure of CaCl₂and urea solutions of the same concentration at the same temperature are respectively 0.605 atm and 0.245 atm. Calculate van’t Hoff factor for CaCl₂.
Answer:
Given : π_Cacl2= 0.605 atm;
π_Urea= 0.245 atm
For urea solution, van’t Hoff factor, i = 1
π_Cacl2= i × (CRT)_Cacl2
π_Urea= (CRT)_Urea

van’t Hoff factor = i = 2.47

Question v.
Explain reverse osmosis.
Answer:
Reverse osmosis : The phenomenon of the passage of solvent like water under high pressure from the concentrated aqueous solution like seawater into pure water through a semipermeable membrane is called reverse osmosis.

The osmotic pressure of seawater is about 30 atmospheres. Hence when pressure more than 30 atmospheres is applied on the solution side, regular osmosis stops and reverse osmosis starts. Hence pure water from seawater enters the other side of pure water.

Purification of seawater by reverse osmosis

For this purpose of suitable semipermeable membrane is required which can withstand high pressure conditions over a long period.
This method is used successfully in Florida since 1981 producing more than 10 million litres of pure water per day.

Question vi.
How molar mass of a solute is determined by osmotic pressure measurement?
Answer:
Consider V dm³(litres) of a solution containing W₂mass of a solute of molar mass M₂at a temperature T.
Number of moles of solute, n₂=
The osmotic pressure π is given by,
π =
∴ M₂=
By measuring osmotic pressure of a solution, the molar mass of a solute can be calculated.
Since osmotic pressure can be measured more precisely, it is widely used to measure molar masses of the substances.

Question vii.
Why vapour pressure of a solvent is lowered by dissolving a nonvolatile solute into it?
Answer:
Lowering of vapour pressure of a solution :
When a nonvolatile solute is added to a pure solvent, the surface area is covered by the solute molecule decreasing the rate of evaporation, hence its vapour pressure decreases. This decrease in vapour pressure is called lowering of vapour pressure.

If P₀is the vapour pressure of a pure solvent (liquid) and P is the vapour pressure of the solution, where P < P₀, then, (P₀– P) is the lowering of the vapour pressure.

Question viii.
Using Raoult’s law, how will you show that ∆P = x₂? Where x₂is the mole fraction of solute in the solution and vapour pressure of pure solvent.
Answer:
If x₁and x₂are the mole fractions of solvent and solute respectively, then
x₁+ x₂
By Raoult’s law,
P = x₁× P₀
where P⁰is the vapour pressure of a pure solvent and P is the vapour pressure of the solution at given temperature.

Question ix.
While considering boiling point elevation and freezing point depression a solution concentration is expressed in molality and not in molarity. Why?
Answer:

At equilibrium,
AxBy ⇌ xA^y++ yB^x-
For 1 mole of electrolyte : 1 – α, xα, yα and
For ‘m’ moles of an electrolyte : m(1 – α), mxα, myα are the number of particles.
Total number of moles at equilibrium, will be,
Total moles = m(1 – α) + mxα + myα
= m[(1 – α) + xα + yα]
= m[1 – xα + yα – α]
= m[1 + α(x + y – 1)]
The van’t Hoff factor i will be,

If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n
∴ i = 1 + α(n – 1)
∴ α(n – 1) = i – 1
∴ α = ……..(1)
This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.

This variation in solubility with temperature can be used to separate the salts from the mixture by fractional crystallisation.

(2) When a liquid is heated, its vapour pressure rises till it becomes equal to the external pressure.
If the liquid has a low vapour pressure, it has a higher boiling point.

Vapour pressure curve showing elevation in boiling point

(3) When a nonvolatile solute is added to a solvent, its vapour pressure decreases, hence the boiling point increases.
This is explained by graphical representation of the vapour pressure and the boiling point of the pure solvent and the solution.

If T₀and T are the boiling points of a pure solvent and a solution, then the elevation in the boiling point is given by,
ΔT_b= T – T₀
The curve AB, represents the variation in the vapour pressure of a pure solvent with temperature while curve CD represents the variation in the vapour pressure of the solution.

(4) This elevation in the boiling point is proportional to the lowering of the vapour pressure, i.e., P₀– P, where P₀and P are the vapour pressures of the pure solvent and the solution.
[ΔT_b∝ (P₀– P) or ΔT_b∝ ΔP]