Answer the following. (i) Boiling point of water at 750 mm Hg pressure is 99.68 °C. How much sucrose (Molar mass = 342 g mol^-1) is to be added to 500 g of water such that it boils at 100 °C ? (K_b for water = 0.52 K kg mol^-1). (ii) State Henry's law and write its any one application.

ChemistryCBSE Class 12 · Board exam 2024, set 56/5/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2024: question 32(b)

Set in the CBSE Class 12 board exam of 2024, paper 56/5/1, section E, question 32(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(b) · 5 marks · Long answerSolutions

Answer the following.

(i)
Boiling point of water at 750 mm Hg pressure is 99.68 °C. How much sucrose (Molar mass = 342 g $mo l^{- 1}$ ) is to be added to 500 g of water such that it boils at 100 °C ? ( $K_{b}$ for water = 0.52 K kg $mo l^{- 1}$ ).
3 marks
(ii)
State Henry's law and write its any one application.
2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 56/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

The water must boil at 100 °C instead of 99.68 °C, so first find the elevation in boiling point.

Hint 2

Use $Δ T_{b} = K_{b} \cdot m$ , where the molality $m = \frac{w _{B} \times 1000}{M _{B} \times w _{A}}$ , and rearrange it for $w_{B}$ .

Hint 3

For Henry's law, link the solubility of a gas to its partial pressure, then think of a fizzy drink bottle.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Rise in boiling point needed: $Δ T_{b} = 100 - 99.68 = 0.32 °C = 0.32 K$
½ mark
Formula: $Δ T_{b} = K_{b} \times \frac{w _{B} \times 1000}{M _{B} \times w _{A}}$ , so $w_{B} = \frac{Δ T _{b} \times M _{B} \times w _{A}}{K _{b} \times 1000}$
1 mark
Substitution: $w_{B} = \frac{0.32 \times 342 \times 500}{0.52 \times 1000} = \frac{0.32 \times 342}{0.52 \times 2} = \frac{109.44}{1.04}$
1 mark
$w_{B} = 105.23 g$ of sucrose must be added.
½ mark
(ii) Henry's law: At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. In mole-fraction form, the partial pressure of the gas in the vapour phase is proportional to its mole fraction in the solution: $p = K_{H} x$ .
1 mark
Application: Bottles of soft drinks and soda water are sealed under high pressure so that more CO $_{2}$ dissolves in them. Other acceptable applications are adding helium to scuba-divers' tanks to avoid the bends, and the low O $_{2}$ in the blood of climbers at high altitude (anoxia).
1 mark

Answer: (i) 105.23 g sucrose; (ii) $p = K_{H} x$ at constant temperature; application: soft-drink bottles are sealed under high CO $_{2}$ pressure.

Where the marks are

Calculates $Δ T_{b} = 0.32$ K (½ mark)
Writes the formula $Δ T_{b} = K_{b} w_{B} \times 1000/ (M_{B} w_{A})$ (1 mark)
Substitutes the values correctly (1 mark)
Gets $w_{B} = 105.23$ g with unit; ½ mark is deducted if the unit is missing or wrong (½ mark)
States Henry's law correctly (1 mark)
Gives any one correct application (1 mark)

Revise the chapter

Chapter: Solutions (notes, formulas, where marks are lost and every past question)
Topic: Elevation of Boiling Point and Depression of Freezing Point

Practise this chapterBoard-style questions from Solutions, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 56/5/1: every question, by section.

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CBSE Class 12 Chemistry 2024: question 32(b)

Hints

Step-by-step solution

Revise the chapter

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