Answer the following. (i) Ishan's automobile radiator is filled with 1.0 kg of water. How many grams of ethylene glycol (Molar mass = 62 g mol^-1) must Ishan add to get the freezing point of the solution lowered to –2.8 °C. K_f for water is 1.86 K kg ·\ mol^-1. (ii) What type of deviation from Raoult's law is shown by ethanol and acetone mixture ? Give reason.

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

CBSE Class 12 Chemistry 2024: question 32(a)

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

Q32(a) · 5 marks · Long answerSolutions

Answer the following.

(i)
Ishan's automobile radiator is filled with 1.0 kg of water. How many grams of ethylene glycol (Molar mass = 62 g $mo l^{- 1}$ ) must Ishan add to get the freezing point of the solution lowered to –2.8 °C. $K_{f}$ for water is 1.86 K kg $\cdot mo l^{- 1}$ .
3 marks
(ii)
What type of deviation from Raoult's law is shown by ethanol and acetone mixture ? Give reason.
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 freezing point of water is 0 °C, so $Δ T_{f}$ = 2.8 K; use the molality form of the freezing point depression formula.

Hint 2

For (ii), compare ethanol–ethanol hydrogen bonding with ethanol–acetone attraction.

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) Depression in freezing point: $Δ T_{f} = 0 - (- 2.8) = 2.8$ K.
Formula: $Δ T_{f} = K_{f} \times \frac{w _{B}}{M _{B} \times W _{A} ( kg )}$
1 mark
Substitute: $2.8 = 1.86 \times \frac{w _{B}}{62 \times 1.0}$
1 mark
$w_{B} = \frac{2.8 \times 62}{1.86} = 93.33$ g of ethylene glycol
1 mark
(ii) An ethanol–acetone mixture shows positive deviation from Raoult's law.
1 mark
Reason: pure ethanol molecules are held by hydrogen bonds. Acetone molecules come between them and break some of these bonds, so ethanol–acetone attractions are weaker than ethanol–ethanol attractions. Molecules escape more easily, and the vapour pressure is higher than Raoult's law predicts ( $Δ_{mi x} H > 0$ ).
1 mark

Answer: (i) 93.33 g ethylene glycol; (ii) Positive deviation – ethanol–acetone interactions weaker than ethanol–ethanol (H-bonds broken)

Where the marks are

$Δ T_{f}$ = 2.8 K and correct formula (1 mark)
Correct substitution (1 mark)
$w_{B}$ = 93.33 g with unit (1 mark; ½ deducted for missing unit)
Positive deviation (1 mark)
Reason: weaker A–B interactions, H-bonds of ethanol broken (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(a)

Hints

Step-by-step solution

Revise the chapter

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