0.5 molal aqueous solution of a weak acid ( HX ) is 20\% ionised. If K_f for water is 1.86 K kg mol^-1 , the lowering in freezing point of the solutio - Chemistry MCQ Chemistry | ExamTest.live

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0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If K_{f} for water is 1.86 K kg mol^{- 1}, the lowering in freezing point of the solution is:

A $- 1.12 K$
B $0.56 K$
C $1.12 K$
D $- 0.56 K$

Solution & Step-by-step Explanation

For HX \to H^{+} + X^{-}, the van't Hoff factor i = 1 + α (n - 1) .Given α = 0.20 (20%) and n = 2 . i = 1 + 0.20 (2 - 1) = 1.20 Lowering in freezing point Δ T_{f} = i \cdot K_{f} \cdot m : Δ T_{f} = 1.20 \times 1.86 \times 0.5 = 1.116 \approx 1.12 K Lowering is a positive value; the freezing point itself would be - 1.12^{\circ} C .

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0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If K_{f} for water is 1.86 K kg mol^{- 1}, the lowering in freezing point of the solution is:

- 1.12 K

0.56 K

1.12 K

- 0.56 K

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$0.5 molal$ aqueous solution of a weak acid ( $HX$ ) is $20%$ ionised. If $K_{f}$ for water is $1.86 K kg mol^{- 1}$ , the lowering in freezing point of the solution is:

Solution & Step-by-step Explanation

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0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf​ for water is 1.86 K kg mol−1, the lowering in freezing point of the solution is:

Solution & Step-by-step Explanation

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$0.5 molal$ aqueous solution of a weak acid ( $HX$ ) is $20%$ ionised. If $K_{f}$ for water is $1.86 K kg mol^{- 1}$ , the lowering in freezing point of the solution is: