Chapter 2: Solutions Mock Test
Time: 1 Hour | Maximum Marks: 25
- All questions are compulsory.
- Section A contains Q1 (Multiple Choice) and Q2 (Very Short Answer).
- Section B contains Short Answer Type I questions (2 marks each). Attempt any 4.
- Section C contains Short Answer Type II questions (3 marks each). Attempt any 2.
- Section D contains Long Answer questions (4 marks each). Attempt any 1.
- Use of logarithmic tables is allowed. Calculators are not permitted.
SECTION A
Q1. Select and write the most appropriate answer from the given alternatives: [4 Marks]
-
Which of the following is NOT a colligative property?
(A) Osmotic pressure(B) Surface tension(C) Elevation in boiling point(D) Depression in freezing point
-
The unit of ebullioscopic constant ($K_b$) is:
(A) K kg mol$^{-1}$(B) K mol kg$^{-1}$(C) K kg$^{-1}$ mol$^{-1}$(D) K mol$^{-1}$
-
Two solutions are isotonic if they have the same:
(A) Vapor pressure(B) Density(C) Osmotic pressure(D) Boiling point
-
The van't Hoff factor ($i$) for a dilute aqueous solution of $Na_2SO_4$ is approximately:
(A) 1(B) 2(C) 3(D) 4
Q2. Answer the following questions in one sentence: [3 Marks]
- State Henry's law.
- What is meant by reverse osmosis?
- Define Ebullioscopic constant.
SECTION B
Attempt any FOUR of the following: [8 Marks]
- Distinguish between ideal and non-ideal solutions. (Any 2 points).
- Explain why the vapor pressure of a solvent lowers when a non-volatile solute is dissolved in it.
- The solubility of $N_2$ gas in water at 298 K and 1 atm is $6.8 \times 10^{-4} \text{ mol L}^{-1}$. Calculate the Henry's law constant.
- What are hypertonic and hypotonic solutions? Explain with an example.
- State Raoult's law for a solution of a non-volatile solute. Write its mathematical expression.
SECTION C
Attempt any TWO of the following: [6 Marks]
- Derive the relationship between the elevation of boiling point and the molar mass of a non-volatile solute.
- A solution containing 0.5126 g of naphthalene (molar mass 128 g/mol) in 50 g of carbon tetrachloride yields a boiling point elevation of 0.402 K. Calculate the molal elevation constant of carbon tetrachloride.
- 5% aqueous solution (by mass) of cane sugar (molar mass 342 g/mol) is isotonic with 0.877% aqueous solution of an unknown solute. Calculate the molar mass of the unknown solute.
SECTION D
Attempt any ONE of the following: [4 Marks]
- (a) The vapor pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea ($NH_2CONH_2$) is dissolved in 850 g of water. Calculate the relative lowering of vapor pressure for this solution. (Molar mass of urea = 60 g/mol, Molar mass of water = 18 g/mol). [3 Marks]
(b) Define: Cryoscopic constant ($K_f$). [1 Mark] - (a) Explain the van't Hoff factor ($i$) and write its relationship with the degree of dissociation ($\alpha$) for a weak electrolyte. [2 Marks]
(b) Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C? ($K_b$ for water = 0.52 K kg/mol, Molar mass of sucrose = 342 g/mol). [2 Marks]
Solutions & Marking Scheme
SECTION A [7 Marks]
Q1. Multiple Choice Answers:
1. (B) Surface tension [1 Mark for correct option]
2. (A) K kg mol$^{-1}$ [1 Mark for correct option]
3. (C) Osmotic pressure [1 Mark for correct option]
4. (C) 3 [1 Mark. $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$, so 3 particles]
Q2. Very Short Answers:
1. Henry's Law:
It states that the solubility ($S$) of a gas in a liquid at a constant temperature is directly proportional to the partial pressure ($P$) of the gas present above the surface of the solution. [1 Mark for correct statement]
2. Reverse Osmosis:
If a pressure larger than the osmotic pressure is applied to the solution side, the pure solvent flows out of the solution through the semipermeable membrane into the pure solvent side. This is called reverse osmosis. [1 Mark for correct definition]
3. Ebullioscopic Constant ($K_b$):
It is defined as the elevation in boiling point produced when exactly 1 mole of a non-volatile, non-electrolyte solute is dissolved in 1 kg (1000 g) of the solvent. [1 Mark for correct definition]
SECTION B [8 Marks]
Q3. Ideal vs Non-ideal solutions:
| Ideal Solutions | Non-ideal Solutions |
|---|---|
| Strictly obey Raoult's law over the entire range of concentrations. | Do not obey Raoult's law. |
| $\Delta H_{mix} = 0$ and $\Delta V_{mix} = 0$. | $\Delta H_{mix} \neq 0$ and $\Delta V_{mix} \neq 0$. |
[1 Mark for each correct point of distinction. Total 2 Marks]
Q4. Lowering of vapor pressure:
Evaporation is a surface phenomenon. In a pure solvent, the entire surface area is occupied by volatile solvent molecules. [1 Mark]
When a non-volatile solute is added, the solute particles occupy some of the surface area, decreasing the number of volatile molecules at the surface. Thus, the rate of evaporation decreases, leading to a lowered vapor pressure. [1 Mark]
Q5. Henry's Law Numerical:
Given: $S = 6.8 \times 10^{-4} \text{ mol L}^{-1}$, $P = 1 \text{ atm}$. [1/2 Mark]
Formula: $S = K_H \cdot P \implies K_H = \frac{S}{P}$ [1/2 Mark]
Calculation: $K_H = \frac{6.8 \times 10^{-4}}{1} = 6.8 \times 10^{-4}$ [1/2 Mark]
Answer: $K_H = 6.8 \times 10^{-4} \text{ mol L}^{-1} \text{ atm}^{-1}$ [1/2 Mark for correct unit]
Q6. Hypertonic and Hypotonic Solutions:
If two solutions have different osmotic pressures, the one with the higher osmotic pressure is called hypertonic, and the one with the lower osmotic pressure is called hypotonic. [1 Mark]
Example: A 0.1 M NaCl solution is hypertonic with respect to a 0.05 M NaCl solution. Therefore, 0.05 M NaCl is hypotonic with respect to 0.1 M NaCl. [1 Mark]
Q7. Raoult's Law for non-volatile solute:
Statement: The vapor pressure of a solvent in a solution containing a non-volatile solute is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution. [1 Mark]
Expression: $P_1 = P_1^\circ \cdot x_1$
(Also accepted: $\frac{P_1^\circ - P_1}{P_1^\circ} = x_2$) [1 Mark]
SECTION C [6 Marks]
Q8. Elevation in BP Derivation:
1. Elevation in BP is proportional to molality: $\Delta T_b = K_b \cdot m$ [1 Mark]
2. Molality $m = \frac{n_2}{W_1 \text{ (in kg)}} = \frac{W_2 / M_2}{W_1 / 1000} = \frac{1000 \cdot W_2}{M_2 \cdot W_1}$ [1 Mark]
3. Substitute $m$: $\Delta T_b = K_b \cdot \left[ \frac{1000 \cdot W_2}{M_2 \cdot W_1} \right]$
4. $M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1}$ [1 Mark]
Q9. Numerical on $K_b$:
Given: $W_2 = 0.5126 \text{ g}$, $M_2 = 128 \text{ g/mol}$, $W_1 = 50 \text{ g}$, $\Delta T_b = 0.402 \text{ K}$. Find $K_b$. [1/2 Mark]
Formula: $M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1} \implies K_b = \frac{M_2 \cdot \Delta T_b \cdot W_1}{1000 \cdot W_2}$ [1/2 Mark]
Calculation: $K_b = \frac{128 \times 0.402 \times 50}{1000 \times 0.5126} = \frac{2572.8}{512.6}$ [1 Mark]
Answer: $K_b = 5.019 \text{ K kg mol}^{-1}$ [1 Mark for answer with units]
Q10. Numerical on Isotonic Solutions:
Given: Isotonic $\implies \pi_1 = \pi_2$. For cane sugar: $W_1 = 5\text{g}$, $V_1 = 100\text{mL}$, $M_1 = 342$. For unknown: $W_2 = 0.877\text{g}$, $V_2 = 100\text{mL}$. [1/2 Mark]
Since $\pi = \frac{W}{MV}RT$, we get $\frac{W_1}{M_1} = \frac{W_2}{M_2}$ (as V, R, T are same). [1 Mark]
$\frac{5}{342} = \frac{0.877}{M_2} \implies M_2 = \frac{0.877 \times 342}{5}$ [1/2 Mark]
Answer: $M_2 = 59.98 \text{ g/mol}$ (approx 60 g/mol). [1 Mark for correct calculation]
SECTION D [4 Marks]
Q11. (a) RLVP Numerical [3 Marks] (b) Cryoscopic Constant [1 Mark]
(a) RLVP Calculation:
Moles of urea $n_2 = 50/60 = 0.833 \text{ mol}$. Moles of water $n_1 = 850/18 = 47.22 \text{ mol}$. [1 Mark]
Mole fraction of solute $x_2 = \frac{n_2}{n_1 + n_2} = \frac{0.833}{47.22 + 0.833} = \frac{0.833}{48.053} = 0.0173$. [1 Mark]
Relative lowering of VP = $x_2 = 0.0173$. [1 Mark]
(b) Cryoscopic constant ($K_f$): It is the depression in freezing point produced when 1 mole of a non-volatile solute is dissolved in 1 kg of solvent. [1 Mark]
Q12. (a) van't Hoff Factor [2 Marks] (b) BP Elevation Numerical [2 Marks]
(a) van't Hoff factor ($i$):
It is the ratio of the observed colligative property to the theoretical colligative property. [1 Mark]
Relationship with degree of dissociation ($\alpha$): $\alpha = \frac{i - 1}{n - 1}$ where $n$ is the number of ions produced from one formula unit. [1 Mark]
(b) BP Numerical:
Given: $T_b^\circ = 99.63^\circ\text{C}$, $T_b = 100^\circ\text{C}$, $W_1 = 500\text{g}$, $K_b = 0.52$, $M_2 = 342$. Find $W_2$.
$\Delta T_b = 100 - 99.63 = 0.37 \text{ K}$. [1/2 Mark]
$\Delta T_b = \frac{1000 \cdot K_b \cdot W_2}{M_2 \cdot W_1} \implies W_2 = \frac{\Delta T_b \cdot M_2 \cdot W_1}{1000 \cdot K_b}$ [1/2 Mark]
$W_2 = \frac{0.37 \times 342 \times 500}{1000 \times 0.52} = \frac{63270}{520}$ [1/2 Mark]
Answer: $W_2 = 121.67 \text{ g}$ [1/2 Mark]
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