JEE Chemistry Solutions MCQs Set 27

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Multiple Choice Questions

Question 1: A solution is made by mixing 1 mol of benzene (\( p^0=100 \) mm Hg) and 1 mol of toluene (\( p^0=40 \) mm Hg). The pressure over the mixture at 300 K is slowly reduced. At what pressure does the first bubble of vapour form?
(a) 70 mm Hg
(b) 140 mm Hg
(c) 90 mm Hg
(d) 65 mm Hg
Show Answer & Explanation

Answer: (a) 70 mm Hg

Explanation:
1. The first bubble forms at the bubble point, where the total vapour pressure equals the external pressure.
2. \( P=0.5\times100+0.5\times40=70 \) mm Hg.

Teacher's Note:
1. At pressures above 70 mm Hg there is no vapour.
2. This is the bubble point curve.

Question 2: A solution is made by mixing 1 mol of benzene (\( p^0=100 \) mm Hg) and 1 mol of toluene (\( p^0=40 \) mm Hg). The pressure is reduced to the bubble point. The composition (mole fraction) of the first bubble of vapour is:
(a) Benzene 2/7, toluene 5/7
(b) Benzene 3/7, toluene 4/7
(c) Benzene 1/7, toluene 6/7
(d) Benzene 5/7, toluene 2/7
Show Answer & Explanation

Answer: (d) Benzene 5/7, toluene 2/7

Explanation:
1. \( p_B=50 \), \( p_T=20 \), total 70 mm Hg.
2. \( y_B=50/70=5/7 \); \( y_T=20/70=2/7 \).

Teacher's Note:
1. The vapour is richer in the more volatile benzene.
2. The liquid is 1/2 : 1/2.

Question 3: A solution is made by mixing 1 mol of benzene (\( p^0=100 \) mm Hg) and 1 mol of toluene (\( p^0=40 \) mm Hg). The pressure at which 1 mol of the mixture (half of the total) has been vaporised is:
(a) 70 mm Hg
(b) 100 mm Hg
(c) 63.25 mm Hg
(d) 40 mm Hg
Show Answer & Explanation

Answer: (c) 63.25 mm Hg

Explanation:
1. Let x be the mole fraction of benzene in the liquid left (1 mol), so the vapour (1 mol) has \( y_B=1-x \).
2. \( P=40+60x \) and \( y_B=\dfrac{100x}{P} \), so \( 100x=(1-x)(40+60x) \) giving \( 3x^2+4x-2=0 \).
3. x = 0.3874 and P = 40 + 60 x 0.3874 = 63.25 mm Hg.

Teacher's Note:
1. The pressure lies between the bubble point (70) and the dew point.
2. The dew point pressure (last drop of liquid) is about 57.1 mm Hg.

Question 4: Match each solute in List I (0.1 mol dissolved in 10 mol of water, degree of dissociation alpha as given) with its characteristic in List II. P. \( \ce{AlCl3} \), alpha = 0.8; Q. \( \ce{BaCl2} \), alpha = 0.9; R. \( \ce{Na3PO4} \), alpha = 0.9; S. \( \ce{K4[Fe(CN)6]} \), alpha = 0.7. List II: 1. i = 3.4; 2. minimum osmotic pressure; 3. minimum freezing point; 4. relative lowering of vapour pressure = 37/1037.
(a) P-1, Q-2, R-4, S-3
(b) P-4, Q-2, R-3, S-1
(c) P-2, Q-1, R-3, S-4
(d) P-1, Q-2, R-3, S-4
Show Answer & Explanation

Answer: (a) P-1, Q-2, R-4, S-3

Explanation:
1. i = 1 + (n-1)alpha: AlCl3 (n = 4): 3.4; BaCl2 (n = 3): 2.8; Na3PO4 (n = 4): 3.7; K4[Fe(CN)6] (n = 5): 3.8.
2. Smallest i (BaCl2) gives minimum osmotic pressure; largest i (K4[Fe(CN)6]) gives minimum freezing point.
3. RLVP = 0.37/10.37 = 37/1037 for i = 3.7, which is Na3PO4.

Teacher's Note:
1. RLVP = n_solute particles / (n_solvent + n_particles).
2. Osmotic pressure and freezing point depression both depend on i.

Question 5: In the depression of freezing point experiment it is found that: I. the vapour pressure of the solution is less than that of pure solvent; II. the vapour pressure of the solution is more than that of pure solvent; III. only solute molecules solidify at the freezing point; IV. only solvent molecules solidify at the freezing point. The correct statements are:
(a) I and II
(b) II and III
(c) I, II and III
(d) I and IV
Show Answer & Explanation

Answer: (d) I and IV

Explanation:
1. A non-volatile solute lowers the vapour pressure of the solvent.
2. At the freezing point only pure solid solvent separates; the solute remains in the solution.

Teacher's Note:
1. The freezing point is where solid solvent and solution have equal vapour pressure.
2. Solute separates only at the eutectic point.

Question 6: The van't Hoff factor for a 0.1 M \( \ce{Ba(NO3)2} \) solution is 2.74. The degree of dissociation is:
(a) 91.3%
(b) 87%
(c) 100%
(d) 74%
Show Answer & Explanation

Answer: (b) 87%

Explanation:
1. Ba(NO3)2 gives 3 ions, so \( i=1+2\alpha \).
2. \( \alpha=\dfrac{2.74-1}{2}=0.87 \), i.e. 87%.

Teacher's Note:
1. Complete dissociation would give i = 3.
2. i = 1 + (n - 1) alpha.

Question 7: 3.0 g of acetic acid is added to 500 cm\(^3\) of water (density 0.997 g cm\(^{-3}\)). If 23% of the acetic acid is dissociated, the depression in freezing point is (\( K_f=1.86 \) K kg mol\(^{-1}\)):
(a) 0.186 K
(b) 0.114 K
(c) 0.229 K
(d) 0.458 K
Show Answer & Explanation

Answer: (c) 0.229 K

Explanation:
1. Moles of acid = 3/60 = 0.05; mass of water = 498.5 g, so m = 0.05/0.4985 = 0.1003.
2. \( i=1+\alpha=1.23 \).
3. \( \Delta T_f=1.23\times1.86\times0.1003=0.229 \) K.

Teacher's Note:
1. Acetic acid is a monobasic weak acid giving two particles.
2. Without dissociation the depression would be 0.186 K.

Question 8: When 20 g of naphthoic acid (\( \ce{C11H8O2} \), M = 172) is dissolved in 50 g of benzene (\( K_f=1.72 \) K kg mol\(^{-1}\)), a freezing point depression of 2 K is observed. The van't Hoff factor i is:
(a) 0.5
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (a) 0.5

Explanation:
1. Molality = (20/172)/0.05 = 2.326 mol/kg.
2. Expected depression for i = 1: 1.72 x 2.326 = 4.0 K.
3. \( i=2/4.0=0.5 \): the acid dimerises in benzene.

Teacher's Note:
1. i < 1 shows association.
2. Carboxylic acids form hydrogen-bonded dimers in non-polar solvents.

Question 9: A solution M is made by mixing ethanol and water with mole fraction of ethanol 0.9. Data: \( K_f \) of ethanol 2.0 K kg/mol, freezing point of ethanol 155.7 K, \( K_b \) of water 0.52 K kg/mol, boiling point of water 373 K, vapour pressure of pure water 32.8 mm Hg and of pure ethanol 40 mm Hg; M(water) = 18, M(ethanol) = 46. Treat the solutions as ideal and dilute, with non-volatile non-dissociating solutes. The vapour pressure of solution M is:
(a) 36.0 mm Hg
(b) 29.5 mm Hg
(c) 39.3 mm Hg
(d) 28.8 mm Hg
Show Answer & Explanation

Answer: (c) 39.3 mm Hg

Explanation:
1. Both ethanol and water are volatile here, so use Raoult's law for each: \( P=0.9\times40+0.1\times32.8 \).
2. = 36 + 3.28 = 39.28, about 39.3 mm Hg.

Teacher's Note:
1. 36.0 would count ethanol alone.
2. Dalton's law: total pressure is the sum of partial pressures.

Question 10: A solution M is made by mixing ethanol and water with mole fraction of ethanol 0.9. Data: \( K_f \) of ethanol 2.0 K kg/mol, freezing point of ethanol 155.7 K, \( K_b \) of water 0.52 K kg/mol, boiling point of water 373 K, vapour pressure of pure water 32.8 mm Hg and of pure ethanol 40 mm Hg; M(water) = 18, M(ethanol) = 46. Treat the solutions as ideal and dilute, with non-volatile non-dissociating solutes. Water is added to M so that the mole fraction of water becomes 0.9. The boiling point of this solution (ethanol now the solute) is:
(a) 376.2 K
(b) 380.4 K
(c) 375.5 K
(d) 354.7 K
Show Answer & Explanation

Answer: (a) 376.2 K

Explanation:
1. Per mole of solution: 0.9 mol water (16.2 g) and 0.1 mol ethanol.
2. Molality of ethanol in water = 0.1/0.0162 = 6.17 mol/kg.
3. \( \Delta T_b=0.52\times6.17=3.21 \); boiling point = 373 + 3.21 = 376.2 K.

Teacher's Note:
1. Solvent is now water: use Kb of water.
2. Ethanol is treated as non-volatile for this question.

Question 11: The Henry's law constant for N\(_2\) in water at 298 K is \( 1.0\times10^5 \) atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 mol of water at 298 K and 5 atm pressure is:
(a) \( 4.0\times10^{-5} \)
(b) \( 5.0\times10^{-4} \)
(c) \( 4.0\times10^{-6} \)
(d) \( 4\times10^{-4} \)
Show Answer & Explanation

Answer: (d) \( 4\times10^{-4} \)

Explanation:
1. Partial pressure of N2 = 0.8 x 5 = 4 atm.
2. \( X_{N_2}=4/10^5=4\times10^{-5} \).
3. \( n=4\times10^{-5}\times10=4\times10^{-4} \) mol.

Teacher's Note:
1. Use the partial pressure.
2. n is negligible against 10 mol of water.

Question 12: The freezing point of a solution containing 0.1 g of \( \ce{K3[Fe(CN)6]} \) (M = 329) in 100 g of water (\( K_f=1.86 \) K kg mol\(^{-1}\)) is:
(a) \( -5.7\times10^{-2}\ ^\circ\text{C} \)
(b) \( -2.3\times10^{-2}\ ^\circ\text{C} \)
(c) \( -5.7\times10^{-3}\ ^\circ\text{C} \)
(d) \( -1.2\times10^{-2}\ ^\circ\text{C} \)
Show Answer & Explanation

Answer: (b) \( -2.3\times10^{-2}\ ^\circ\text{C} \)

Explanation:
1. m = (0.1/329)/0.1 = 3.04 x 10^-3 mol/kg.
2. The complex gives 3 K+ and [Fe(CN)6]3-: i = 4.
3. \( \Delta T_f=4\times1.86\times3.04\times10^{-3}=2.26\times10^{-2} \) K.

Teacher's Note:
1. Without dissociation it is 5.7 x 10^-3 K.
2. Use i = 4.

Question 13: For a dilute solution of 2.5 g of a non-volatile non-electrolyte in 100 g of water, the elevation in boiling point at 1 atm is 2 C (\( K_b=0.76 \) K kg mol\(^{-1}\)). Assuming the solute concentration is much lower than that of the solvent, the vapour pressure (mm Hg) of the solution is (normal boiling point pressure 760 mm Hg):
(a) 740
(b) 736
(c) 724
(d) 718
Show Answer & Explanation

Answer: (c) 724

Explanation:
1. m = 2/0.76 = 2.632 mol/kg, so the moles of solute in 100 g = 0.2632.
2. Moles of water = 5.556; for a dilute solution \( X_2\approx n_2/n_1=0.04737 \).
3. \( p=760(1-0.04737)=724 \) mm Hg.

Teacher's Note:
1. The exact mole fraction gives 725.6, the same as 724 within the approximation.
2. Boiling point elevation is related to the lowering of vapour pressure.

Question 14: \( \ce{MX2} \) dissociates into \( \ce{M^{2+}} \) and \( \ce{X-} \) in aqueous solution with degree of dissociation 0.5. The ratio of the observed depression of freezing point to the value without dissociation is:
(a) 1.5
(b) 2
(c) 3
(d) 0.5
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. \( i=1+(3-1)\times0.5=2 \).
2. Ratio of observed to calculated depression = i = 2.

Teacher's Note:
1. Complete dissociation would give i = 3.
2. MX2 gives 3 ions.

Question 15: The freezing point of a 0.01 molal aqueous solution of a cobalt(III) chloride-ammonia complex (a strong electrolyte) is -0.0558 C (\( K_f=1.86 \)). The number of chloride ions in the coordination sphere of the complex is:
(a) 1
(b) 0
(c) 2
(d) 3
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. \( i=\dfrac{0.0558}{1.86\times0.01}=3 \).
2. Three particles: one complex ion and two free Cl-, so the complex is \( \ce{[Co(NH3)5Cl]Cl2} \).
3. One chloride is in the coordination sphere.

Teacher's Note:
1. Total Cl for Co(III) is 3.
2. i = 4 would be [Co(NH3)6]Cl3 (no chloride inside).

Question 16: Which of the following mixtures show positive deviation from Raoult's law at 35 C? (1) carbon tetrachloride + methanol, (2) carbon disulphide + acetone, (3) benzene + toluene, (4) phenol + aniline.
(a) (1) and (3)
(b) (2) and (4)
(c) (3) and (4)
(d) (1) and (2)
Show Answer & Explanation

Answer: (d) (1) and (2)

Explanation:
1. Methanol with CCl4 and CS2 with acetone: the hydrogen bonding or dipole interactions of the polar liquid are broken, so A-B attraction is weaker: positive deviation.
2. Benzene + toluene is nearly ideal; phenol + aniline form hydrogen bonds and show negative deviation.

Teacher's Note:
1. Carbon disulphide is non-polar and acetone polar.
2. Negative deviations: acetone + chloroform, phenol + aniline.

Question 17: On dissolving 0.5 g of a non-volatile non-ionic solute in 39 g of benzene, its vapour pressure decreases from 650 to 640 mm Hg. The depression of freezing point of benzene (K) is: (benzene M = 78, \( K_f=5.12 \) K kg mol\(^{-1}\))
(a) 1.02 K
(b) 0.51 K
(c) 2.04 K
(d) 0.25 K
Show Answer & Explanation

Answer: (a) 1.02 K

Explanation:
1. \( X_2=10/650=0.01538 \); moles of benzene = 39/78 = 0.5.
2. \( n_2=\dfrac{X_2n_1}{1-X_2}=0.00781 \) mol, so molality = 0.00781/0.039 = 0.200.
3. \( \Delta T_f=5.12\times0.2003=1.02 \) K.

Teacher's Note:
1. Solute mass is not needed for the answer.
2. The molar mass of the solute would be 64 g/mol.

Question 18: Benzene (\( p^0=75 \) torr) and toluene (\( p^0=22 \) torr) form a nearly ideal solution at 20 C. The partial vapour pressure of benzene for a solution of 78 g benzene and 46 g toluene is:
(a) 25 torr
(b) 50 torr
(c) 37.5 torr
(d) 53.5 torr
Show Answer & Explanation

Answer: (b) 50 torr

Explanation:
1. Moles: benzene 78/78 = 1; toluene 46/92 = 0.5.
2. \( X_B=1/1.5=0.667 \).
3. \( p_B=0.667\times75=50 \) torr.

Teacher's Note:
1. Partial pressure uses the mole fraction in the liquid.
2. Total pressure is 50 + 7.3 = 57.3 torr.

Question 19: A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm Hg at 300 K. The vapour pressure of propyl alcohol is 200 mm Hg. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure at the same temperature is:
(a) 700 mm Hg
(b) 360 mm Hg
(c) 350 mm Hg
(d) 300 mm Hg
Show Answer & Explanation

Answer: (c) 350 mm Hg

Explanation:
1. \( 290=0.6\,p_E^0+0.4\times200 \).
2. \( p_E^0=(290-80)/0.6=350 \) mm Hg.

Teacher's Note:
1. Ideal solution of two similar alcohols.
2. The question asks for the pure-liquid vapour pressure of ethanol.

Question 20: A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (M = 60 g/mol) in the same solvent. The densities of both solutions are assumed equal to 1.0 g cm\(^{-3}\). The molar mass of the substance is:
(a) 105 g/mol
(b) 210 g/mol
(c) 90 g/mol
(d) 15 g/mol
Show Answer & Explanation

Answer: (b) 210 g/mol

Explanation:
1. Isotonic: equal molarity, so \( \dfrac{5.25}{M}=\dfrac{1.5}{60} \).
2. \( M=\dfrac{5.25\times60}{1.5}=210 \) g/mol.

Teacher's Note:
1. The substance is assumed to be a non-electrolyte.
2. Equal densities make % w/v equal to % w/w numerically.

Question 21: The vapour pressure of water at 20 C is 17.5 mm Hg. If 18 g of glucose is added to 178.2 g of water at 20 C, the vapour pressure of the solution is:
(a) 17.325 mm Hg
(b) 15.750 mm Hg
(c) 16.500 mm Hg
(d) 17.675 mm Hg
Show Answer & Explanation

Answer: (a) 17.325 mm Hg

Explanation:
1. Moles of glucose = 18/180 = 0.1; moles of water = 178.2/18 = 9.9.
2. \( X_{water}=9.9/10=0.99 \).
3. \( p=0.99\times17.5=17.325 \) mm Hg.

Teacher's Note:
1. Glucose is non-volatile.
2. The relative lowering equals X_glucose = 0.01.

Question 22: At 80 C the vapour pressure of pure liquid A is 520 mm Hg and that of pure B is 1000 mm Hg. A solution of A and B boils at 80 C and 1 atm (760 mm Hg). The amount of A in the mixture is:
(a) 34 mol per cent
(b) 48 mol per cent
(c) 52 mol per cent
(d) 50 mol per cent
Show Answer & Explanation

Answer: (d) 50 mol per cent

Explanation:
1. At the boiling point the total vapour pressure is 760 mm Hg.
2. \( 520x_A+1000(1-x_A)=760 \), so \( x_A=240/480=0.5 \).

Teacher's Note:
1. Assume an ideal solution.
2. 760 mm Hg = 1 atm.

Question 23: A binary liquid solution is made by mixing n-heptane and ethanol. Which statement is correct?
(a) The solution is non-ideal, showing negative deviation
(b) n-Heptane shows positive deviation while ethanol shows negative deviation
(c) The solution is non-ideal, showing positive deviation from Raoult's law
(d) The solution is ideal
Show Answer & Explanation

Answer: (c) The solution is non-ideal, showing positive deviation from Raoult's law

Explanation:
1. Ethanol molecules are strongly hydrogen bonded; the non-polar heptane breaks these bonds without giving comparable attraction.
2. So the vapour pressure is higher than ideal (positive deviation for both components).

Teacher's Note:
1. Both components deviate in the same direction.
2. Delta H mix is positive.

Question 24: Two liquids X and Y form an ideal solution. At 300 K, the vapour pressure of a solution of 1 mol X and 3 mol Y is 550 mm Hg. If 1 mol of Y is added, the vapour pressure increases by 10 mm Hg. The vapour pressures of pure X and pure Y are, respectively:
(a) 400 and 600 mm Hg
(b) 300 and 400 mm Hg
(c) 500 and 600 mm Hg
(d) 200 and 300 mm Hg
Show Answer & Explanation

Answer: (a) 400 and 600 mm Hg

Explanation:
1. \( \dfrac{p_X+3p_Y}{4}=550 \), so \( p_X+3p_Y=2200 \).
2. \( \dfrac{p_X+4p_Y}{5}=560 \), so \( p_X+4p_Y=2800 \).
3. Subtract: \( p_Y=600 \), \( p_X=400 \) mm Hg.

Teacher's Note:
1. Check: (400 + 1800)/4 = 550.
2. Two equations for two unknowns.

Question 25: If sodium sulphate is completely dissociated, the change in freezing point of water when 0.01 mol of \( \ce{Na2SO4} \) is dissolved in 1 kg of water is (\( K_f=1.86 \) K kg mol\(^{-1}\)):
(a) 0.0372 K
(b) 0.0744 K
(c) 0.0186 K
(d) 0.0558 K
Show Answer & Explanation

Answer: (d) 0.0558 K

Explanation:
1. i = 3 for complete dissociation.
2. \( \Delta T_f=3\times1.86\times0.01=0.0558 \) K.

Teacher's Note:
1. 0.0186 K is the value without dissociation.
2. Na2SO4 gives 2 Na+ and 1 SO4(2-).

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