JEE Chemistry Solutions MCQs Set 24

Multiple Choice Questions (MCQs) for JEE Chemistry: Solutions

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Multiple Choice Questions

Question 1: The osmotic pressures at 273 K of solutions of 10 g glucose (P1), 10 g urea (P2) and 10 g sucrose (P3), each in 250 mL of water, are related as:
(a) \( P_1>P_2>P_3 \)
(b) \( P_3>P_1>P_2 \)
(c) \( P_2>P_1>P_3 \)
(d) \( P_2>P_3>P_1 \)
Show Answer & Explanation

Answer: (c) \( P_2>P_1>P_3 \)

Explanation:
1. Same mass, same volume and temperature, so \( \pi\propto \) moles \( \propto 1/M \).
2. Moles: urea 10/60 = 0.167; glucose 10/180 = 0.0556; sucrose 10/342 = 0.0292.
3. Order: urea > glucose > sucrose.

Teacher's Note:
1. The lighter the molecule, the more moles in 10 g.
2. All three are non-electrolytes.

Question 2: The osmotic pressure of blood is 7.40 atm at body temperature (37 degrees C). The number of moles of glucose needed per litre for an intravenous injection that has the same osmotic pressure as blood is (R = 0.0821 L atm/mol K):
(a) 0.29, i.e. about 0.3
(b) 0.2
(c) 0.1
(d) 0.4
Show Answer & Explanation

Answer: (a) 0.29, i.e. about 0.3

Explanation:
1. \( C=\dfrac{\pi}{RT}=\dfrac{7.40}{0.0821\times310}=0.29 \) mol/L.

Teacher's Note:
1. Isotonic solutions have equal osmotic pressure at the same temperature.
2. A 5% glucose solution (0.28 M) is the common isotonic drip.

Question 3: The total concentration of dissolved particles inside red blood cells is about 0.30 M. If the cells are placed in pure water at 298 K, the osmotic pressure inside the cells becomes (R = 0.0821 L atm/mol K):
(a) 1.78 atm
(b) 2.34 atm
(c) 0.74 atm
(d) 7.34 atm
Show Answer & Explanation

Answer: (d) 7.34 atm

Explanation:
1. \( \pi=CRT=0.30\times0.0821\times298=7.34 \) atm.

Teacher's Note:
1. Water enters the cell because the pure water outside is hypotonic; the cell may burst.
2. Use total particle concentration directly.

Question 4: A 17.4% (w/v) \( \ce{K2SO4} \) solution at 27 degrees C is isotonic with a 5.85% (w/v) NaCl solution at 27 degrees C. If NaCl is 100% ionised, the percentage ionisation of \( \ce{K2SO4} \) is (M of K2SO4 = 174):
(a) 25%
(b) 50%
(c) 75%
(d) 100%
Show Answer & Explanation

Answer: (b) 50%

Explanation:
1. K2SO4: 17.4 g/100 mL = 1.0 M. NaCl: 5.85 g/100 mL = 1.0 M, i = 2.
2. Isotonic means \( i_1C_1=i_2C_2 \Rightarrow i_{K_2SO_4}=2 \).
3. \( i=1+2\alpha=2 \Rightarrow \alpha=0.5 \), i.e. 50%.

Teacher's Note:
1. Isotonic solutions have equal total particle concentration.
2. K2SO4 gives 3 ions on complete ionisation.

Question 5: At 2 degrees C the osmotic pressure of a solution is 500 mm Hg. The solution is diluted and the temperature raised to 27 degrees C, when the osmotic pressure is 109.09 mm Hg. The solution was diluted to:
(a) 2 times its original volume
(b) 4 times its original volume
(c) 5 times its original volume
(d) 10 times its original volume
Show Answer & Explanation

Answer: (c) 5 times its original volume

Explanation:
1. \( \pi\propto C\,T \). \( C_1\propto \dfrac{500}{275} \), \( C_2\propto\dfrac{109.09}{300} \).
2. \( \dfrac{C_2}{C_1}=\dfrac{0.3636}{1.818}=0.2 \), so the volume is multiplied by 5.

Teacher's Note:
1. Use absolute temperatures: 275 K and 300 K.
2. Concentration is inversely proportional to volume.

Question 6: A 0.20 M NaCl solution is completely evaporated and 0.150 mol of NaCl is recovered. The original volume of the sample was:
(a) 30 mL
(b) 750 mL
(c) 333 mL
(d) 1000 mL
Show Answer & Explanation

Answer: (b) 750 mL

Explanation:
1. \( V=n/M=0.150/0.20=0.75 \) L = 750 mL.

Teacher's Note:
1. Moles of solute stay the same on evaporation.
2. Convert litres to millilitres at the end.

Question 7: A 20.0 mL sample of \( \ce{CuSO4} \) solution is evaporated to dryness leaving 0.967 g of residue. The molarity of the original solution is (Cu = 63.5):
(a) 0.303 M
(b) 0.0484 M
(c) 0.0207 M
(d) 48.4 M
Show Answer & Explanation

Answer: (a) 0.303 M

Explanation:
1. Molar mass of CuSO4 = 63.5 + 32 + 64 = 159.5 g/mol.
2. Moles \( =0.967/159.5=6.06\times10^{-3} \).
3. \( M=6.06\times10^{-3}/0.020=0.303 \) M.

Teacher's Note:
1. Convert 20.0 mL to 0.0200 L.
2. The residue is the anhydrous salt.

Question 8: The vapour pressure of water at 20 degrees C is 17.54 mm Hg. If the piston in a closed vessel containing liquid water and its vapour is lowered, halving the volume above the liquid at constant temperature, the vapour pressure of water becomes:
(a) 8.77 mm Hg
(b) 35.08 mm Hg
(c) Between 8.77 and 17.54 mm Hg
(d) 17.54 mm Hg
Show Answer & Explanation

Answer: (d) 17.54 mm Hg

Explanation:
1. Vapour pressure depends only on the liquid and temperature.
2. Some vapour condenses to keep the equilibrium pressure unchanged.

Teacher's Note:
1. Gas laws apply only to a gas without a liquid reservoir.
2. Volume and surface area of the container do not change the vapour pressure.

Question 9: A sample of air saturated with benzene (vapour pressure 100 mm Hg at 298 K) is at 298 K and a total pressure of 750 mm Hg. If it is isothermally compressed to one-third of its initial volume, the final total pressure is:
(a) 2250 torr
(b) 2150 torr
(c) 2050 torr
(d) 1950 torr
Show Answer & Explanation

Answer: (c) 2050 torr

Explanation:
1. Benzene vapour stays at its saturation pressure of 100 torr (the excess condenses).
2. Air: \( 650\times3=1950 \) torr (Boyle's law).
3. Total \( =1950+100=2050 \) torr.

Teacher's Note:
1. Saturated vapour does not obey Boyle's law while liquid is present.
2. If benzene also behaved as a gas the answer would be 2250 torr.

Question 10: Water and chlorobenzene are immiscible. Their mixture boils at 89 degrees C under a reduced pressure of \( 7.7\times10^4 \) Pa. The vapour pressure of pure water at 89 degrees C is \( 7\times10^4 \) Pa. The weight percent of chlorobenzene (M = 112.5) in the distillate is:
(a) 38.46
(b) 50
(c) 60
(d) 78.3
Show Answer & Explanation

Answer: (a) 38.46

Explanation:
1. Boiling: \( p_{water}+p_{CB}=7.7\times10^4 \Rightarrow p_{CB}=0.7\times10^4 \) Pa.
2. \( \dfrac{w_{CB}}{w_W}=\dfrac{p_{CB}M_{CB}}{p_WM_W}=\dfrac{0.7\times112.5}{7\times18}=0.625 \).
3. Percent \( =\dfrac{0.625}{1.625}\times100=38.46 \).

Teacher's Note:
1. Steam distillation: mass ratio in distillate = (p M) ratio.
2. The distillate separates into two layers.

Question 11: Two completely immiscible liquids A (pure vapour pressure 100 torr) and B (200 torr) are present in a closed vessel. The total vapour pressure of the system is:
(a) Less than 100 torr
(b) Between 100 and 200 torr
(c) Less than 200 torr
(d) 300 torr
Show Answer & Explanation

Answer: (d) 300 torr

Explanation:
1. Each liquid behaves independently, so \( P=p^0_A+p^0_B=100+200=300 \) torr.

Teacher's Note:
1. Immiscible liquids boil at a temperature below the boiling point of either.
2. Miscible ideal mixtures give a value between the two pure values.

Question 12: Pure liquids A and B have vapour pressures 300 and 800 torr at 25 degrees C. When mixed to form a solution with 92 mole percent B, the total vapour pressure is 0.95 atm. Which is true for the solution?
(a) \( \Delta V_{mix}>0 \)
(b) \( \Delta H_{mix}<0 \)
(c) \( \Delta V_{mix}=0 \)
(d) \( \Delta H_{mix}=0 \)
Show Answer & Explanation

Answer: (b) \( \Delta H_{mix}<0 \)

Explanation:
1. Ideal value \( =0.08\times300+0.92\times800=760 \) torr = 1.0 atm.
2. Observed \( 0.95\) atm \( =722 \) torr, lower than ideal: negative deviation.
3. Negative deviation means \( \Delta H_{mix}<0 \) and \( \Delta V_{mix}<0 \).

Teacher's Note:
1. Convert atm to torr: 0.95 x 760 = 722.
2. Positive deviation would give positive enthalpy and volume change.

Question 13: Barium ions, cyanide ions and \( \ce{Co^{2+}} \) form an ionic complex. If the complex is 75% ionised in water with a van't Hoff factor of 4, the coordination number of Co(II) in the complex can be:
(a) 6
(b) 4
(c) 5
(d) 6 and 4 both
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. Let the complex be \( \ce{Ba_y[Co(CN)_x]_z} \) with charge balance \( 2y=z(x-2) \). It gives \( y+z \) ions.
2. \( i=1+\alpha(n-1)=4 \Rightarrow 0.75(n-1)=3 \Rightarrow n=5 \), so \( y+z=5 \).
3. Trying integers: z = 2, y = 3 gives \( 6=2(x-2) \Rightarrow x=5 \). No other combination gives an integer x. The complex is \( \ce{Ba3[Co(CN)5]2} \).

Teacher's Note:
1. The charge of \( \ce{[Co(CN)_x]} \) is \( 2-x \).
2. The general expression is \( i=1+\alpha(n-1) \) for n ions.

Question 14: Among 1 m sucrose, 1 m potassium ferricyanide and 1 m potassium sulphate, the solution with the maximum vapour pressure is:
(a) 1 m potassium ferricyanide
(b) 1 m sucrose
(c) 1 m potassium sulphate
(d) All equal
Show Answer & Explanation

Answer: (b) 1 m sucrose

Explanation:
1. Vapour pressure is lowered most by the solution with most particles: ferricyanide (4 ions) > sulphate (3 ions) > sucrose (1 particle).
2. So sucrose gives the highest vapour pressure.

Teacher's Note:
1. Colligative effects depend on particle number.
2. Assume complete ionisation of the salts.

Question 15: x moles of KCl and y moles of \( \ce{BaCl2} \) are dissolved in 1 kg of water, with \( x+y=0.1 \). If \( K_f \) for water is 1.85 K/molal, the observed range of \( \Delta T_f \) as the ratio x : y is varied is:
(a) 0.37 to 0.555 degrees C
(b) 0.185 to 0.93 degrees C
(c) 0.56 to 0.93 degrees C
(d) 0.37 to 0.93 degrees C
Show Answer & Explanation

Answer: (a) 0.37 to 0.555 degrees C

Explanation:
1. \( \Delta T_f=1.85(2x+3y) \) with \( y=0.1-x \).
2. All KCl: \( 1.85\times0.2=0.37 \). All BaCl2: \( 1.85\times0.3=0.555 \).
3. The expression is linear, so the range is 0.37 to 0.555.

Teacher's Note:
1. Extremes of a linear function occur at the ends of the allowed range.
2. KCl gives 2 ions and BaCl2 gives 3.

Question 16: Two beakers, one with 20 mL of 0.05 M non-volatile non-electrolyte solution and the other with 20 mL of 0.03 M NaCl solution (assume complete dissociation), are placed side by side in a closed enclosure. The volumes in the first and second beaker at equilibrium are:
(a) 21.8 mL and 18.2 mL
(b) 20 mL and 20 mL
(c) 17.1 mL and 22.9 mL
(d) 18.2 mL and 21.8 mL
Show Answer & Explanation

Answer: (d) 18.2 mL and 21.8 mL

Explanation:
1. Effective concentrations: 0.05 M and 0.06 M. Solvent transfers via the vapour from the lower to the higher concentration until they are equal.
2. Moles of particles: 0.001 and 0.0012 in a total volume of 40 mL.
3. \( V_1:V_2=1:1.2 \Rightarrow V_1=18.2 \) mL, \( V_2=21.8 \) mL.

Teacher's Note:
1. The solution with the lower vapour pressure (NaCl) gains water.
2. At equilibrium both have equal particle concentration.

Question 17: What volume (mL) of 98% sulphuric acid (density 1.88 g/mL) must be mixed with water to make 200 mL of 15% (by weight) solution of density 1.12 g/mL?
(a) 33.6 mL
(b) 16.2 mL
(c) 18.2 mL
(d) 22.4 mL
Show Answer & Explanation

Answer: (c) 18.2 mL

Explanation:
1. Mass of 200 mL solution \( =224 \) g; \( \ce{H2SO4} \) needed \( =0.15\times224=33.6 \) g.
2. Mass of 98% acid \( =33.6/0.98=34.29 \) g.
3. Volume \( =34.29/1.88=18.2 \) mL.

Teacher's Note:
1. 33.6 is a mass of pure acid, not a volume.
2. Densities convert between mass and volume.

Question 18: An ideal aqueous solution contains liquid A (M = 128) at 64% by weight. Its vapour pressure is 145 mm Hg and that of pure water at the same temperature is 155 mm Hg. If the vapour pressure of pure A is x mm Hg, then x/5 is:
(a) 21
(b) 25
(c) 105
(d) 29
Show Answer & Explanation

Answer: (a) 21

Explanation:
1. In 100 g: A = 64 g = 0.5 mol, water = 36 g = 2 mol. \( x_A=0.2,\ x_W=0.8 \).
2. \( 145=0.2\,p^0_A+0.8\times155 \Rightarrow 0.2p^0_A=21 \Rightarrow p^0_A=105 \).
3. \( x/5=21 \).

Teacher's Note:
1. 105 is x itself, not x/5.
2. Both components contribute to the total vapour pressure.

Question 19: A and B form ideal solutions. At 50 degrees C, \( p^0_A \) is half \( p^0_B \). A solution of 0.2 mol A and 0.8 mol B has a normal boiling point of 50 degrees C. If \( p^0_B \) is in atm, the value of \( 18\,p^0_B \) is:
(a) 10
(b) 18
(c) 36
(d) 20
Show Answer & Explanation

Answer: (d) 20

Explanation:
1. Normal boiling point: total pressure 1 atm.
2. \( 0.2\left(\dfrac{p^0_B}{2}\right)+0.8p^0_B=0.9p^0_B=1 \Rightarrow p^0_B=\dfrac{10}{9} \) atm.
3. \( 18\times\dfrac{10}{9}=20 \).

Teacher's Note:
1. Use the mole fractions 0.2 and 0.8 directly as the numbers of moles add to 1.
2. Normal boiling means 1 atm.

Question 20: The vapour pressure of pure A at 300 K is 577 torr and that of pure B is 390 torr. They form ideal liquid and gaseous mixtures. In a mixture in which the mole fraction of A in the vapour is 0.35, the mole percent of A in the liquid is closest to:
(a) 35.0
(b) 26.7
(c) 40.0
(d) 12.5
Show Answer & Explanation

Answer: (b) 26.7

Explanation:
1. \( y_A=\dfrac{x_A\,577}{x_A\,577+(1-x_A)\,390}=0.35 \).
2. \( 577x_A=0.35(390+187x_A) \Rightarrow 511.55x_A=136.5 \Rightarrow x_A=0.267 \).

Teacher's Note:
1. A is more volatile, so it is richer in the vapour than in the liquid; the liquid contains less A than 35%.
2. Check: 0.267 x 577 = 154; total = 154 + 0.733 x 390 = 440; 154/440 = 0.35.

Question 21: How many of the following facts about a chloroform and ethyl methyl ketone solution are correct? (a) It shows negative deviation. (b) It forms a maximum boiling azeotropic mixture. (c) \( \Delta S>0 \). (d) \( \Delta G<0 \). (e) The components can be separated by fractional distillation.
(a) 3
(b) 5
(c) 4
(d) 2
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Hydrogen bonding between CHCl3 and the ketone gives negative deviation: (a) true.
2. Negative deviation gives a maximum boiling azeotrope: (b) true. Mixing raises entropy and is spontaneous: (c), (d) true.
3. An azeotrope cannot be separated by fractional distillation: (e) false. So 4 are correct.

Teacher's Note:
1. Azeotropes boil at a constant temperature with fixed composition.
2. Positive deviation gives a minimum boiling azeotrope.

Question 22: The solubility of a gas in a liquid:
(a) Increases with increase in both pressure and temperature
(b) Increases with increase in pressure and decreases with increase in temperature
(c) Decreases with increase in pressure and temperature
(d) Decreases with pressure and increases with temperature
Show Answer & Explanation

Answer: (b) Increases with increase in pressure and decreases with increase in temperature

Explanation:
1. Henry's law: solubility is proportional to partial pressure.
2. Dissolution of a gas is exothermic, so by Le Chatelier's principle solubility falls on heating.

Teacher's Note:
1. Warm water holds less dissolved oxygen.
2. Soda water fizzes when the pressure is released.

Question 23: The solubility of \( \ce{N2} \) in water exposed to the atmosphere at a partial pressure of 593 mm is \( 5.3\times10^{-4} \) M. Its solubility at 760 mm and the same temperature is:
(a) \( 6.8\times10^{-4} \) M
(b) \( 4.1\times10^{-4} \) M
(c) 1500 M
(d) 2400 M
Show Answer & Explanation

Answer: (a) \( 6.8\times10^{-4} \) M

Explanation:
1. Henry's law: solubility proportional to pressure.
2. \( S=5.3\times10^{-4}\times\dfrac{760}{593}=6.8\times10^{-4} \) M.

Teacher's Note:
1. Higher pressure gives higher solubility, ruling out the smaller value.
2. Do not multiply the two pressures.

Question 24: A liquid immiscible with water was steam distilled at 95.2 degrees C at a total pressure of 748 torr. The distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water at 95.2 degrees C is 648 torr. The molar mass of the liquid is:
(a) 7.975 g/mol
(b) 166 g/mol
(c) 18 g/mol
(d) 145.8 g/mol
Show Answer & Explanation

Answer: (d) 145.8 g/mol

Explanation:
1. Vapour pressure of the liquid \( =748-648=100 \) torr.
2. \( \dfrac{w_L}{w_W}=\dfrac{p_LM_L}{p_WM_W} \Rightarrow 1.25=\dfrac{100M_L}{648\times18} \).
3. \( M_L=145.8 \) g/mol.

Teacher's Note:
1. Total pressure at boiling is the sum of the two pure vapour pressures.
2. Use the mass ratio, not the mole ratio.

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