JEE Chemistry Solutions MCQs Set 23

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Multiple Choice Questions

Question 1: An ideal solution contains two volatile liquids A (pure vapour pressure 100 torr) and B (200 torr). A mixture of 1 mole of A and 4 moles of B is boiled and the vapour is condensed completely. The total vapour pressure of this condensate (distillate) at the same temperature is:
(a) 150 torr
(b) 180 torr
(c) 188.9 torr
(d) 198.9 torr
Show Answer & Explanation

Answer: (c) 188.9 torr

Explanation:
1. Liquid: \( x_A=0.2,\ x_B=0.8 \). \( p_A=0.2\times100=20 \) torr, \( p_B=0.8\times200=160 \) torr, total 180 torr.
2. Vapour composition: \( y_A=20/180=1/9,\ y_B=8/9 \). The condensate has this composition.
3. \( P=\tfrac19(100)+\tfrac89(200)=188.9 \) torr.

Teacher's Note:
1. The vapour is richer in the more volatile component (B), so the distillate has a higher vapour pressure than the original liquid (180 torr).
2. Repeating distillation raises it further towards 200 torr.

Question 2: At 323 K the vapour pressure (in mm Hg) of a methanol-ethanol solution is given by \( p=120X_A+140 \), where \( X_A \) is the mole fraction of methanol. The value of \( \lim_{X_A\to1}\dfrac{p_A}{X_A} \) is:
(a) 260 mm
(b) 250 mm
(c) 140 mm
(d) 120 mm
Show Answer & Explanation

Answer: (a) 260 mm

Explanation:
1. For an ideal mixture \( p=p^0_AX_A+p^0_B(1-X_A)=p^0_B+(p^0_A-p^0_B)X_A \).
2. Comparing: \( p^0_B=140 \), \( p^0_A-p^0_B=120 \), so \( p^0_A=260 \).
3. As \( X_A\to1 \), Raoult's law holds for A and \( p_A/X_A\to p^0_A=260 \) mm.

Teacher's Note:
1. The intercept at \( X_A=0 \) is the vapour pressure of pure B.
2. The value at \( X_A=1 \) is the sum 120 + 140 = 260.

Question 3: The vapour pressures of pure A and B at 350 K are 300 torr and 800 torr. For an ideal mixture of A and B with a normal boiling point of 350 K (760 torr), the mole fraction of A is:
(a) 0.06
(b) 0.04
(c) 0.02
(d) 0.08
Show Answer & Explanation

Answer: (d) 0.08

Explanation:
1. Normal boiling point: total vapour pressure = 760 torr.
2. \( 300x_A+800(1-x_A)=760 \Rightarrow 500x_A=40 \Rightarrow x_A=0.08 \).

Teacher's Note:
1. Normal boiling point means boiling under 1 atm = 760 torr.
2. B is far more volatile, so the liquid is mostly B.

Question 4: Two liquids A and B have pure vapour pressures in the ratio 1 : 3 and the ratio of the numbers of moles of A and B in the liquid phase is 1 : 3. The mole fraction of A in the vapour phase in equilibrium is:
(a) 0.2
(b) 0.1
(c) 0.5
(d) 1.0
Show Answer & Explanation

Answer: (b) 0.1

Explanation:
1. Take \( p^0_A=1,\ p^0_B=3 \), \( x_A=1/4,\ x_B=3/4 \).
2. \( p_A=1/4,\ p_B=9/4 \), total \( =10/4 \).
3. \( y_A=\dfrac{1/4}{10/4}=0.1 \).

Teacher's Note:
1. \( y_A = p_A/p_{total} \) from Dalton's law.
2. B is three times more volatile and three times more abundant, so it dominates the vapour.

Question 5: The vapour pressure of a solution of two liquids A (pure 80 mm) and B (pure 120 mm) is found to be 100 mm when \( x_A=0.4 \). The result shows that the solution:
(a) Is ideal
(b) Shows positive deviation from Raoult's law
(c) Shows negative deviation from Raoult's law
(d) Shows positive deviation at low concentration and negative at high concentration
Show Answer & Explanation

Answer: (c) Shows negative deviation from Raoult's law

Explanation:
1. Ideal value: \( 0.4\times80+0.6\times120=32+72=104 \) mm.
2. Observed 100 mm is lower than 104 mm, so there is negative deviation.

Teacher's Note:
1. Negative deviation: A-B attractions stronger than A-A and B-B, \( \Delta H_{mix}<0 \).
2. Positive deviation gives a vapour pressure higher than the ideal value.

Question 6: For a binary mixture of volatile liquids with \( X_A=0.4 \), the vapour pressure is 580 torr (\( p^0_A=300 \), \( p^0_B=800 \) torr). The mixture could be:
(a) \( \ce{C6H5Cl} \) and \( \ce{C6H5Br} \)
(b) \( \ce{CHCl3} \) and \( \ce{CH3COCH3} \)
(c) \( \ce{C6H6} \) and \( \ce{C6H5CH3} \)
(d) n-hexane and n-heptane
Show Answer & Explanation

Answer: (b) \( \ce{CHCl3} \) and \( \ce{CH3COCH3} \)

Explanation:
1. Ideal value: \( 0.4\times300+0.6\times800=120+480=600 \) torr. The observed 580 torr is lower than 600 torr, i.e. negative deviation.

Teacher's Note:
1. Chloroform and acetone form hydrogen bonds with each other, giving negative deviation.
2. The other three pairs are nearly ideal.

Question 7: When KCl dissolves in water the process is endothermic. Which set of signs is correct for the dissolution?
(a) \( \Delta H=+ve,\ \Delta S=+ve,\ \Delta G=-ve \)
(b) \( \Delta H=+ve,\ \Delta S=+ve,\ \Delta G=+ve \)
(c) \( \Delta H=+ve,\ \Delta S=-ve,\ \Delta G=-ve \)
(d) \( \Delta H=-ve,\ \Delta S=-ve,\ \Delta G=-ve \)
Show Answer & Explanation

Answer: (a) \( \Delta H=+ve,\ \Delta S=+ve,\ \Delta G=-ve \)

Explanation:
1. Dissolution is spontaneous, so \( \Delta G<0 \).
2. Endothermic: \( \Delta H>0 \). The crystal breaks into mobile ions, so disorder increases: \( \Delta S>0 \).
3. \( \Delta G=\Delta H-T\Delta S<0 \) because the entropy term outweighs the enthalpy term.

Teacher's Note:
1. Spontaneous endothermic processes are driven by entropy.
2. Raising the temperature increases the solubility of KCl.

Question 8: One mole of a solute A associates in solution as \( \ce{nA <=> A_n} \). If \( \alpha \) is the degree of association, the van't Hoff factor is:
(a) \( i=1-\alpha \)
(b) \( i=1+\dfrac{\alpha}{n} \)
(c) \( i=1 \)
(d) \( i=1-\alpha+\dfrac{\alpha}{n} \)
Show Answer & Explanation

Answer: (d) \( i=1-\alpha+\dfrac{\alpha}{n} \)

Explanation:
1. Initially 1 mol of A. At equilibrium \( (1-\alpha) \) mol of A and \( \alpha/n \) mol of \( A_n \).
2. Total moles \( =1-\alpha+\alpha/n \); \( i=\dfrac{\text{total after}}{\text{initial}} \).

Teacher's Note:
1. For dimerisation (n = 2) this gives \( i=1-\alpha/2 \).
2. Association makes i less than 1.

Question 9: If \( M_{normal} \) is the normal molar mass and \( \alpha \) is the degree of ionisation of \( \ce{K3[Fe(CN)6]} \), the abnormal molar mass of the complex in solution will be:
(a) \( M_{normal}(1+2\alpha)^{-1} \)
(b) \( M_{normal}(1+\alpha)^{-1} \)
(c) \( M_{normal}(1+3\alpha)^{-1} \)
(d) Equal to \( M_{normal} \)
Show Answer & Explanation

Answer: (c) \( M_{normal}(1+3\alpha)^{-1} \)

Explanation:
1. \( \ce{K3[Fe(CN)6] -> 3K+ + [Fe(CN)6]^{3-}} \) gives 4 ions: \( i=1+3\alpha \).
2. Observed molar mass = normal molar mass / i.

Teacher's Note:
1. Colligative effect is proportional to the number of particles; a larger count means a smaller apparent molar mass.
2. For complete ionisation i = 4.

Question 10: A complex containing \( \ce{K+} \), Pt(IV) and \( \ce{Cl-} \) is 100% ionised giving a van't Hoff factor of 3. The complex is:
(a) \( \ce{K2[PtCl6]} \)
(b) \( \ce{K2[PtCl4]} \)
(c) \( \ce{K3[PtCl5]} \)
(d) \( \ce{K[PtCl3]} \)
Show Answer & Explanation

Answer: (a) \( \ce{K2[PtCl6]} \)

Explanation:
1. Pt(IV) with 6 chloride ligands gives \( \ce{[PtCl6]^{2-}} \), which needs two K+.
2. \( \ce{K2[PtCl6] -> 2K+ + [PtCl6]^{2-}} \): 3 ions, i = 3.

Teacher's Note:
1. Check the oxidation state: 2 x (+1) + x - 6 = 0 gives x = +4.
2. \( \ce{K2[PtCl4]} \) contains Pt(II) and \( \ce{K3[PtCl5]} \) contains Pt(III).

Question 11: The relative decrease in vapour pressure of an aqueous NaCl solution containing NaCl in 180 g of water is 0.167. The number of moles of NaCl present is (assume complete dissociation):
(a) 2 mol
(b) 3 mol
(c) 4 mol
(d) 1 mol
Show Answer & Explanation

Answer: (d) 1 mol

Explanation:
1. Moles of water \( N=180/18=10 \). NaCl gives \( i=2 \).
2. \( \dfrac{p^0-p}{p^0}=\dfrac{2n}{2n+10}=\dfrac16 \Rightarrow 12n=2n+10 \Rightarrow n=1 \).

Teacher's Note:
1. 0.167 is 1/6.
2. Always count particles (ions), not formula units, in Raoult's law for electrolytes.

Question 12: The vapour pressure of pure benzene at 50 degrees C is 268 torr. How many moles of non-volatile solute per mole of benzene are needed to lower the vapour pressure to 167 torr?
(a) 0.377
(b) 0.605
(c) 0.623
(d) 0.395
Show Answer & Explanation

Answer: (b) 0.605

Explanation:
1. \( x_{solute}=\dfrac{268-167}{268}=0.377 \).
2. Moles of solute per mole of benzene \( =\dfrac{x_{solute}}{1-x_{solute}}=\dfrac{0.377}{0.623}=0.605 \).

Teacher's Note:
1. 0.377 is the mole fraction of solute, not the mole ratio.
2. The ratio is n / N = (p0 - p) / p.

Question 13: The relative lowering of vapour pressure is 0.4 for a solution containing 1 mol NaCl in 3 mol water. The percentage ionisation of NaCl is:
(a) 60%
(b) 50%
(c) 100%
(d) 40%
Show Answer & Explanation

Answer: (c) 100%

Explanation:
1. \( \dfrac{i\times1}{i\times1+3}=0.4 \Rightarrow 0.6i=1.2 \Rightarrow i=2 \).
2. For NaCl \( i=1+\alpha \), so \( \alpha=1 \): 100% ionisation.

Teacher's Note:
1. Maximum i for NaCl is 2.
2. Use RLVP = i n / (i n + N) for a dissociating solute.

Question 14: The vapour pressure of a solution of a non-volatile solute B in a solvent A is 95% of that of the pure solvent. If the molecular mass of the solvent is 0.3 times that of the solute, the ratio of the mass of solvent to the mass of solute is:
(a) 0.15
(b) 5.7
(c) 0.2
(d) 19
Show Answer & Explanation

Answer: (b) 5.7

Explanation:
1. \( p/p^0=0.95=\dfrac{N}{N+n} \Rightarrow n/N=\dfrac{0.05}{0.95}=\dfrac1{19} \).
2. \( \dfrac{w_A}{w_B}=\dfrac{N M_A}{n M_B}=19\times0.3=5.7 \).

Teacher's Note:
1. Mass = moles x molar mass for each component.
2. The ratio of moles is solvent : solute = 19 : 1.

Question 15: Twenty grams of a non-volatile solute are dissolved in 100 g of water at 25 degrees C. The vapour pressure of pure water is 23.76 mm Hg and that of the solution is 22.41 mm Hg. The molar mass of the solute is about:
(a) 60 g/mol
(b) 30 g/mol
(c) 120 g/mol
(d) 180 g/mol
Show Answer & Explanation

Answer: (a) 60 g/mol

Explanation:
1. \( x_{solute}=\dfrac{23.76-22.41}{23.76}=0.0568 \); \( n/N=0.0568/0.9432=0.0602 \).
2. \( N=100/18=5.556 \) mol, so \( n=0.3347 \) mol.
3. \( M=20/0.3347=59.8\approx60 \) g/mol.

Teacher's Note:
1. The solute is probably urea (M = 60).
2. Mass needed to halve the vapour pressure would be n = N, i.e. 5.556 x 60 = 333 g.

Question 16: The degree of dissociation of \( \ce{Ca(NO3)2} \) in a dilute aqueous solution containing 7 g of the salt per 100 g of water at 100 degrees C is 70%. The vapour pressure of the solution is (vapour pressure of water at 100 degrees C = 760 mm):
(a) 760 mm
(b) 724 mm
(c) 738 mm
(d) 746 mm
Show Answer & Explanation

Answer: (d) 746 mm

Explanation:
1. Moles of salt \( =7/164=0.0427 \); \( i=1+2(0.7)=2.4 \); effective particles \( =0.1024 \) mol.
2. Moles of water \( =100/18=5.556 \).
3. \( p=760\times\dfrac{5.556}{5.556+0.1024}=746.2 \) mm.

Teacher's Note:
1. Ca(NO3)2 gives 3 ions, so i = 1 + 2 alpha.
2. Vapour pressure of water at its normal boiling point is 760 mm.

Question 17: An aqueous solution containing 1 g of urea (M = 60) boils at 100.25 degrees C. An aqueous solution containing 3 g of glucose (M = 180) in the same quantity of water will boil at:
(a) 100.75 degrees C
(b) 100.5 degrees C
(c) 100.25 degrees C
(d) 100 degrees C
Show Answer & Explanation

Answer: (c) 100.25 degrees C

Explanation:
1. Urea: \( 1/60=0.0167 \) mol. Glucose: \( 3/180=0.0167 \) mol.
2. Equal moles of non-electrolytes in the same solvent mass give the same elevation, 0.25 degrees.

Teacher's Note:
1. Elevation depends on the number of particles, not their identity.
2. Both solutes are non-electrolytes, i = 1.

Question 18: The boiling point elevation was 0.52 degrees C when 6 g of a compound X was dissolved in 100 g of water (\( K_b=0.52 \) K kg/mol). The molar mass of X is:
(a) 60 g/mol
(b) 120 g/mol
(c) 100 g/mol
(d) 342 g/mol
Show Answer & Explanation

Answer: (a) 60 g/mol

Explanation:
1. \( m=\Delta T_b/K_b=1 \) mol/kg.
2. Moles in 0.1 kg of water \( =0.1 \); \( M=6/0.1=60 \).

Teacher's Note:
1. \( M=\dfrac{K_b\times1000\,w}{\Delta T_b\times W} \) with W in g.
2. Urea has M = 60.

Question 19: A solute S undergoes reversible trimerisation in a solvent. The boiling point elevation of its 0.1 molal solution equals that of a 0.08 molal solution of a solute which neither associates nor dissociates. The percentage of trimerisation of S is:
(a) 40%
(b) 50%
(c) 60%
(d) 30%
Show Answer & Explanation

Answer: (d) 30%

Explanation:
1. \( \ce{3S <=> S3} \): \( i=1-\alpha+\alpha/3=1-\tfrac{2\alpha}{3} \).
2. \( i\times0.1=0.08 \Rightarrow i=0.8 \).
3. \( \tfrac{2\alpha}{3}=0.2 \Rightarrow \alpha=0.3 \).

Teacher's Note:
1. Equal elevation with the same solvent means equal effective molality.
2. Association reduces the particle count, so i < 1.

Question 20: A complex of iron and cyanide ions is 100% ionised at 1 molal. If its elevation in boiling point is 2.08 degrees C (\( K_b=0.52 \) K kg/mol), the complex is:
(a) \( \ce{Fe(CN)2} \)
(b) \( \ce{K3[Fe(CN)6]} \)
(c) \( \ce{K4[Fe(CN)6]} \)
(d) \( \ce{Fe(CN)4} \)
Show Answer & Explanation

Answer: (b) \( \ce{K3[Fe(CN)6]} \)

Explanation:
1. \( i=\dfrac{\Delta T_b}{K_bm}=\dfrac{2.08}{0.52}=4 \): four ions per formula unit.
2. \( \ce{K3[Fe(CN)6] -> 3K+ + [Fe(CN)6]^{3-}} \) gives 4 ions.
3. \( \ce{K4[Fe(CN)6]} \) would give 5 ions, i = 5.

Teacher's Note:
1. Count the cations plus the complex anion.
2. Fe(CN)2 and Fe(CN)4 are not ionic complexes.

Question 21: \( \ce{PtCl4.6H2O} \) exists as a hydrated complex. Its 1 molal aqueous solution has a depression in freezing point of 3.72 degrees C (\( K_f=1.86 \); assume 100% ionisation). The complex is:
(a) \( \ce{[Pt(H2O)6]Cl4} \)
(b) \( \ce{[Pt(H2O)4Cl2]Cl2.2H2O} \)
(c) \( \ce{[Pt(H2O)3Cl3]Cl.3H2O} \)
(d) \( \ce{[Pt(H2O)2Cl4].4H2O} \)
Show Answer & Explanation

Answer: (c) \( \ce{[Pt(H2O)3Cl3]Cl.3H2O} \)

Explanation:
1. \( i=3.72/1.86=2 \): two particles per formula unit.
2. \( \ce{[Pt(H2O)3Cl3]+Cl-} \) gives 2 ions.
3. The other complexes give 5, 3 and 1 ions.

Teacher's Note:
1. Only ionisable chloride and the complex ion count.
2. Water of crystallisation does not add particles.

Question 22: How many moles of sucrose should be dissolved in 500 g of water so that the difference between the boiling point and the freezing point of the solution is 104 degrees C? (\( K_f=1.86 \), \( K_b=0.52 \) K kg/mol)
(a) 1.68
(b) 0.84
(c) 3.36
(d) 8.40
Show Answer & Explanation

Answer: (b) 0.84

Explanation:
1. Boiling point \( =100+0.52m \); freezing point \( =0-1.86m \).
2. Difference \( =100+(0.52+1.86)m=104 \Rightarrow m=4/2.38=1.68 \) mol/kg.
3. Moles in 0.5 kg water \( =1.68\times0.5=0.84 \).

Teacher's Note:
1. 1.68 is the molality, not the number of moles in 500 g.
2. Sucrose is a non-electrolyte, i = 1.

Question 23: A solution of sucrose in 100 g of water freezes at -0.2 degrees C. As ice separates out, the freezing point falls to -0.25 degrees C. How many grams of ice have separated?
(a) 20 g
(b) 18 g
(c) 25 g
(d) 23 g
Show Answer & Explanation

Answer: (a) 20 g

Explanation:
1. \( \Delta T_f\propto \dfrac{\text{moles solute}}{\text{mass of solvent}} \) and the solute amount is fixed.
2. \( 0.2\times100=0.25\times W \Rightarrow W=80 \) g of water remain.
3. Ice separated \( =100-80=20 \) g.

Teacher's Note:
1. As ice leaves, the solution becomes more concentrated.
2. Use the ratio of depressions rather than computing Kf and moles.

Question 24: Sea water contains 5.85% NaCl and 9.50% \( \ce{MgCl2} \) by weight of solution. Assuming 80% ionisation of NaCl and 50% of \( \ce{MgCl2} \), the normal boiling point of sea water is (\( K_b \) of water = 0.51 K kg/mol):
(a) 101.9 degrees C
(b) 108.5 degrees C
(c) 110.3 degrees C
(d) 102.3 degrees C
Show Answer & Explanation

Answer: (d) 102.3 degrees C

Explanation:
1. In 100 g of solution: NaCl 5.85 g = 0.1 mol, MgCl2 9.5 g = 0.1 mol, water = 84.65 g.
2. \( i_{NaCl}=1+0.8=1.8 \); \( i_{MgCl_2}=1+2(0.5)=2 \). Particles \( =0.18+0.20=0.38 \) mol.
3. \( m=0.38/0.08465=4.49 \); \( \Delta T_b=0.51\times4.49=2.29 \); boiling point \( =102.3 \) degrees C.

Teacher's Note:
1. Percentages are of the solution, so subtract the solutes to get the mass of water.
2. MgCl2 gives 3 ions, i = 1 + 2 alpha.

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