Class 12 Chemistry Chapter 1: Solutions worksheet with answer key
A 20-mark worksheet on Solutions for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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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.
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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.
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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.
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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.
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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.
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(a) Brine is more concentrated than the cell fluid of the mango, so water leaves the mango cells by osmosis and it shrivels. (b) 0.9% NaCl (normal saline) is isotonic with blood-cell fluid, so no osmosis occurs; pure water is hypotonic and would flow into the cells and make them swell.
- water leaves the mango cells by osmosis into the more concentrated brine (1)
- normal saline is isotonic with blood-cell fluid, so cells neither swell nor shrink (1)
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An azeotrope is a binary mixture with the same composition in liquid and vapour, boiling at a constant temperature. Ethanol–water shows a large positive deviation and forms a minimum boiling azeotrope at about 95% ethanol by volume; once this composition is reached, liquid and vapour have the same composition and no further separation occurs.
- definition of azeotrope (1)
- minimum boiling azeotrope at about 95% ethanol stops separation (1)
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The bottle is sealed under high CO₂ pressure. On opening, the partial pressure of CO₂ above the liquid falls; by Henry's law (p = KH x) the solubility falls, and the extra gas escapes as bubbles. In the warm bottle KH is larger, so even less CO₂ stays dissolved at the lower pressure, and more escapes.
- fall in partial pressure of CO₂ lowers its solubility (Henry's law) (1)
- higher KH at higher temperature, so lower solubility and more fizz (1)
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(a) Equal moles in the vapour means pX = pY: 90 xX = 30 (1 − xX), so xX = 0.25. ptotal = 22.5 + 22.5 = 45 kPa. (b) pX = 45 kPa, pY = 15 kPa, ptotal = 60 kPa; yX = 45/60 = 0.75. The vapour is richer in X, the more volatile component.
- pX = pY set up and xX = 0.25 (1)
- ptotal = 45 kPa (1)
- yX = 0.75, vapour richer in the more volatile X (1)
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M₂ = (1000 × w₂ × Kb)/(ΔTb × w₁) = (1000 × 1.80 × 1.20)/(0.240 × 50.0) = 180 g mol⁻¹. Empirical formula mass of CH₂O = 30, so n = 180/30 = 6 and the molecular formula is C₆H₁₂O₆.
- correct expression for M₂ (1)
- M₂ = 180 g mol⁻¹ (1)
- n = 6, molecular formula C₆H₁₂O₆ (1)
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(a) Sucrose: 34.2 g L⁻¹ ÷ 342 = 0.100 M. Isotonic solutions of non-electrolytes at the same temperature have the same molarity, so urea = 0.100 × 60 = 6.0 g L⁻¹ = 0.60% (m/V). (b) Π = CRT = 0.100 × 0.083 × 300 = 2.49 bar. (c) No: NaCl gives two ions, i ≈ 2, so its osmotic pressure is about twice as large.
- 0.60% urea by equal molarity (1)
- Π = 2.49 bar (1)
- NaCl not isotonic, since i ≈ 2 doubles Π (1)