JEE Chemistry Solutions MCQs Set 31

Answer: (c) 0.05 M \( \ce{KNO3} \) > 0.04 M \( \ce{BaCl2} \) > 0.140 M sucrose > 0.075 M \( \ce{CuSO4} \)

Explanation:
1. The greater the total particle concentration (i x M), the lower the freezing point.
2. i x M: \( \ce{KNO3} \) = 2(0.05) = 0.10; \( \ce{BaCl2} \) = 3(0.04) = 0.12; sucrose = 0.140; \( \ce{CuSO4} \) = 2(0.075) = 0.15.
3. Decreasing freezing point means increasing particle concentration: KNO3 > BaCl2 > sucrose > CuSO4.

Teacher's Note:
1. Compare i x concentration for each solute.

Answer: (a) 49

Explanation:
1. \( \ce{BaCl2 -> Ba^{2+} + 2Cl-} \) gives \( i=1+2\alpha \).
2. \( 1.98=1+2\alpha \), so \( \alpha=0.49 \), that is 49%.

Teacher's Note:
1. Each formula unit gives 3 ions, so i = 1 + 2 alpha.

Answer: (d) 10 V

Explanation:
1. Moles of \( \ce{H2O2} \) = \( \frac{6.8}{34}=0.2 \); molarity = \( \frac{0.2}{0.224}=0.893 \) M.
2. Volume strength = 11.2 x molarity = \( 11.2\times0.893=10 \), so the label is 10 V.

Teacher's Note:
1. Volume strength is the litres of O2 at STP given by 1 litre of solution.

Answer: (b) \( S_1 \) is hypertonic while \( S_2 \) is hypotonic

Explanation:
1. Osmotically active concentration of \( S_1 \): \( 2\times0.1=0.20 \) M.
2. For \( S_2 \): \( 3\times0.05=0.15 \) M.
3. \( S_1 \) has the higher osmotic pressure, so it is hypertonic and \( S_2 \) is hypotonic.

Teacher's Note:
1. Hypertonic means higher effective concentration.

Answer: (b) 141.93 mm

Explanation:
1. Moles of solute = \( \frac{0.5}{65}=0.00769 \); mass of \( \ce{CCl4} \) = 158 g, moles = \( \frac{158}{154}=1.026 \).
2. \( x_{solute}=\frac{0.00769}{1.0337}=0.00744 \).
3. \( P=143(1-0.00744)=141.93 \) mm.

Teacher's Note:
1. Raoult's law: P = P0 x (mole fraction of solvent).

Answer: (d) 1% NaCl in water

Explanation:
1. For equal mass percent, the number of particles per 100 g is (1/M) x i.
2. Glucose: 1/180 = 0.0056; sucrose: 1/342 = 0.0029; NaCl: 2/58.5 = 0.0342; \( \ce{CaCl2} \): 3/111 = 0.0270.
3. NaCl gives the most particles, hence the highest boiling point.

Teacher's Note:
1. Small molar mass and large i give more particles.

Answer: (a) 138, 258

Explanation:
1. For an ideal mixture, \( P=P_B^0+(P_A^0-P_B^0)X_A \). Comparing, \( P_B^0=138 \) and \( P_A^0-P_B^0=120 \), so \( P_A^0=258 \).
2. As \( X_A\to0 \), \( P_B/X_B\to P_B^0=138 \); as \( X_B\to0 \), \( P_A/X_A\to P_A^0=258 \).

Teacher's Note:
1. Raoult's law gives P/X = P0 for the nearly pure component.

Answer: (c) 0.5

Explanation:
1. Molar mass = 172 g per mol, so moles = \( \frac{20}{172}=0.1163 \) and molality = \( \frac{0.1163}{0.05}=2.326 \).
2. Expected \( \Delta T_f=1.72\times2.326=4.0 \) K.
3. \( i=\frac{2}{4}=0.5 \) (association into dimers).

Teacher's Note:
1. i less than 1 indicates association.

Answer: (c) 0.2

Explanation:
1. \( x_A=\frac13 \), \( x_B=\frac23 \); take \( P_A^0=1 \), \( P_B^0=2 \).
2. \( p_A=\frac13 \), \( p_B=\frac43 \); total = \( \frac53 \).
3. \( y_A=\frac{1/3}{5/3}=0.2 \).

Teacher's Note:
1. Vapour composition = partial pressure / total pressure.

Answer: (a) 2.99

Explanation:
1. Mass of ethanol = \( 150\times0.78=117 \) g; moles = \( \frac{117}{46}=2.543 \).
2. Water added = 850 mL = 850 g = 0.85 kg.
3. Molality = \( \frac{2.543}{0.85}=2.99 \).

Teacher's Note:
1. Molality uses the mass of the solvent in kg.

Answer: (d) 0.11 RT

Explanation:
1. \( [H^+]=0.01 \) M, so \( \alpha=0.1 \) and \( i=1+\alpha=1.1 \).
2. \( \pi=iCRT=1.1\times0.1\,RT=0.11\,RT \).

Teacher's Note:
1. Include the van't Hoff factor for a weak electrolyte.

Answer: (b) 2.4 atm

Explanation:
1. Urea: \( \frac{1.8}{60}=0.03 \) mol; sucrose: \( \frac{10.26}{342}=0.03 \) mol in the 600 cm\(^3\) mixture.
2. Total concentration = \( \frac{0.06}{0.6}=0.1 \) M.
3. \( \pi=CRT=0.1\times0.0821\times293=2.4 \) atm.

Teacher's Note:
1. Add the moles of non-electrolytes and divide by the final volume.

Answer: (b) \( N_{HCl} \)

Explanation:
1. HCl: 36.59 g per L divided by 36.5 g per equivalent is 1.0 N.
2. Acetic acid: 45.09 g per L divided by 60 g per equivalent is 0.75 N.
3. HCl has the higher normality.

Teacher's Note:
1. Normality = (g per L) / equivalent mass.

Answer: (d) 1.344 M

Explanation:
1. Moles = \( 0.48\times1.5+0.52\times1.2=0.72+0.624=1.344 \).
2. Total volume = 1.0 L, so molarity = 1.344 M.

Teacher's Note:
1. Add moles, then divide by the total volume.

Answer: (a) \( M_A=2M_B \)

Explanation:
1. For a dilute solution, relative lowering = \( \frac{n_{solute}}{n_{solvent}}=\frac{w/m}{W/M} \).
2. With the same w, m and W, it is proportional to the solvent molar mass M.
3. So \( \frac{M_A}{M_B}=2 \).

Teacher's Note:
1. Relative lowering is proportional to the solvent's molar mass.

Answer: (c) 8.21

Explanation:
1. \( \pi=\frac{C}{M}RT \), so \( \log\pi=\log C+\log\frac{RT}{M} \).
2. Intercept = \( \log\frac{0.0821\times300}{300}=\log0.0821=\log(8.21\times10^{-2}) \), so x = 8.21.

Teacher's Note:
1. Intercept occurs where log C = 0.

Answer: (a) 746.27 mm Hg

Explanation:
1. Moles of salt = \( \frac{7}{164}=0.0427 \); \( i=1+2(0.7)=2.4 \), so effective moles = 0.1024.
2. Moles of water = \( \frac{100}{18}=5.556 \); \( x_{solute}=\frac{0.1024}{5.658}=0.0181 \).
3. At 100 \(^\circ\)C, \( P^0=760 \) mm, so \( P=760(1-0.0181)=746.3 \) mm Hg.

Teacher's Note:
1. Pure water boils at 100 C, so its vapour pressure is 760 mm Hg.

Answer: (d) \( 2.96\times10^{-3} \)

Explanation:
1. \( i=\frac{0.06}{1.86\times0.025}=1.29 \), so \( \alpha=0.29 \).
2. \( K_a=\frac{C\alpha^2}{1-\alpha}=\frac{0.025\times0.0841}{0.71}\approx2.96\times10^{-3} \).

Teacher's Note:
1. For a monobasic acid, i = 1 + alpha.