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Class 12 Chemistry Chapter 1: Solutions MCQs with answers

10 multiple-choice questions on Solutions, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
In which of the following is a liquid the solute and a solid the solvent?
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Answer: (d) amalgam of mercury with sodium

In sodium amalgam the liquid mercury is the solute in solid sodium (Table 1.1). Hydrogen in palladium is gas in solid, copper in gold solid in solid, camphor in nitrogen solid in gas.

Question 2
9.2 g of ethanol (C₂H₅OH, 46 g mol⁻¹) is mixed with 36 g of water (18 g mol⁻¹). The mole fraction of ethanol is
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Answer: (b) 0.091

n(ethanol) = 9.2/46 = 0.20 mol; n(water) = 36/18 = 2.0 mol. x = 0.20/(0.20 + 2.0) = 0.091. 0.100 is n₂/n₁, 0.204 the mass fraction, 0.909 the mole fraction of water.

Question 3
2.8 g of KOH (56 g mol⁻¹) is dissolved in water and the solution is made up to 250 mL. Its molarity is
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Answer: (c) 0.20 M

Moles of KOH = 2.8/56 = 0.050 mol; volume = 0.250 L; M = 0.050/0.250 = 0.20 M. 0.050 M forgets the volume; 0.0112 divides grams by millilitres.

Question 4
9.0 g of glucose (C₆H₁₂O₆, 180 g mol⁻¹) is dissolved in 250 g of water. The molality of the solution is
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Answer: (d) 0.20 mol kg⁻¹

Moles = 9.0/180 = 0.050 mol; mass of solvent = 0.250 kg; m = 0.050/0.250 = 0.20 mol kg⁻¹. 0.00020 leaves the solvent in grams.

Question 5
A solution is prepared at 298 K and then warmed to 323 K, with no loss of solvent. Which of its concentrations changes?
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Answer: (d) molarity

Molarity is moles per litre of solution, and volume changes with temperature. Molality, mole fraction and mass percentage are built from masses and moles only, so they are independent of temperature.

Question 6
Which of the following dissolves readily in benzene but hardly at all in water?
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Answer: (b) anthracene

Like dissolves like: non-polar anthracene dissolves in non-polar benzene. NaCl, KCl and sugar are polar or ionic and dissolve in water, not benzene.

Question 7
For a salt X, Δsol H = +18 kJ mol⁻¹. A nearly saturated solution of X is in contact with excess solid. When the temperature is raised, the solubility of X
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Answer: (d) increases, since the equilibrium shifts in the endothermic direction

Δsol H > 0, so dissolution absorbs heat. By Le Chatelier's principle, raising the temperature favours the endothermic change, dissolution, and solubility rises. Incompressibility is the reason pressure, not temperature, has little effect.

Question 8
A gas has a Henry's law constant of 50.0 kbar in water at 298 K. Its mole fraction in water in equilibrium with the gas at a partial pressure of 0.50 bar is
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Answer: (a) 1.0 × 10⁻⁵

p = KH x, so x = p/KH = 0.50 bar / 50 000 bar = 1.0 × 10⁻⁵. 1.0 × 10⁻² leaves KH in kbar; 1.0 × 10⁵ inverts the ratio; 2.5 × 10⁴ multiplies.

Question 9
At 293 K, KH in water is 69.16 kbar for H₂ and 144.97 kbar for He. If each gas is kept over water at the same partial pressure, the ratio x(H₂) : x(He) of their mole fractions in water is about
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Answer: (d) 2.10

x = p/KH, so at equal p, x(H₂)/x(He) = KH(He)/KH(H₂) = 144.97/69.16 = 2.10. The higher KH, the lower the solubility; 0.48 inverts the ratio.

Question 10
KH for O₂ in water is 34.86 kbar at 293 K and 46.82 kbar at 303 K. At the same partial pressure of O₂, the ratio of its mole fraction in water at 293 K to that at 303 K is
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Answer: (c) 1.34

x = p/KH, so x(293 K)/x(303 K) = 46.82/34.86 = 1.34. Less oxygen dissolves in warmer water, which is why aquatic species prefer cold water. 1.03 is the ratio of the temperatures.

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