ChemistryCBSE Class 12 · Board exam 2026, set 56/5/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2026: question 31(a)

Set in the CBSE Class 12 board exam of 2026, paper 56/5/1, section E, question 31(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(a) · 5 marks · Long answerSolutions

(i) Calculate the freezing point of a solution when 10·5 g of $MgB r_{2}$ was dissolved in 250 g of water, assuming $MgB r_{2}$ undergoes complete dissociation. (3)
(Given : Molar mass of $MgB r_{2}$ = 184 g mol $^{- 1}$ , $K_{f}$ for water = 1·86 K kg mol $^{- 1}$ )
(ii) Write two differences between ideal and non-ideal solutions. (1+1)

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 56/5/1.

Hints

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

Hint 1

Count the ions formed when one MgBr₂ unit dissociates completely to get $i$ .

Hint 2

Use $Δ T_{f} = i K_{f} \frac{w _{B} \times 1000}{M _{B} w _{A}}$ , then subtract from the freezing point of water.

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) $MgB r_{2} \to M g^{2 +} + 2B r^{-}$ , so $i = 3$ .
½ mark
$Δ T_{f} = i K_{f} m = i K_{f} \times \frac{w _{B} \times 1000}{M _{B} \times w _{A}}$
½ mark
$Δ T_{f} = \frac{3 \times 1.86 \times 10.5 \times 1000}{184 \times 250} = 1.27 K$
1 mark
Freezing point of solution $= T_{f}^{\circ} - Δ T_{f} = 273.15 - 1.27 = 271.88 K$ (i.e. $- 1.2 7^{\circ}$ C)
1 mark
(ii) Difference 1: an ideal solution obeys Raoult's law over the entire range of concentration, whereas a non-ideal solution does not.
1 mark
Difference 2: for an ideal solution $Δ_{mi x} H = 0$ and $Δ_{mi x} V = 0$ , whereas for a non-ideal solution $Δ_{mi x} H \neq = 0$ and $Δ_{mi x} V \neq = 0$ . (In ideal solutions A–B interactions equal A–A and B–B interactions; in non-ideal solutions they differ.)
1 mark

Answer: (i) Freezing point = 271.88 K (−1.27 °C); (ii) ideal obeys Raoult's law, ΔmixH = ΔmixV = 0; non-ideal does not, ΔmixH, ΔmixV ≠ 0

Where the marks are

i = 3 (½ mark)
Correct formula for ΔTf (½ mark)
ΔTf = 1.27 K (1 mark)
Freezing point 271.88 K (1 mark)
Two correct differences (1 + 1 marks)

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 paper2026 board paper, set 56/5/1: every question, by section.

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CBSE Class 12 Chemistry 2026: question 31(a)

Hints

Step-by-step solution

Revise the chapter

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