$0.5$ 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

0.56 K

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1.12 K

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- 0.56 K

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- 1.12 K

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Solution

The correct option is B $1.12 K$
$\begin{matrix} H X & ⇌ & H^{+} & X^{-} 1 & 0 & 0 1 - α & α & α \end{matrix}$
$∴ i = 1 - α + α + α = 1 + α = 1 + 0.2 = 1.2$
So, the lowering of freezing point of the solution,
$Δ T_{f} = i \times K_{f} \times m = 1.2 \times 1.86 \times 0.5 = 1.116 K \approx 1.12 K$

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0.5 molal aqueous solution of a weak acid $H X$ is 20% ionized. If $K_{f}$ for water is 1.86 $K k g m o l^{- 1}$ , the lowering in freezing point of the solution is

A 0.2 molal aqueous solution of a weak acid (HX) is 20% ionised. The freezing point of this solution is (Given for water)

Q. A

0.2 m o l a l

aqueous solution of a weak acid

H X

20 %

ionized. The freezing point of the solutions is :

[

k_{f} = 1.86 k k g m o l e^{- 1}

for water]

Q. A

0.1

molal aqueous solution of a week acid is

30

% ionised. If

K_{f}

for water is

{1.86}^{\circ} C / m

, the freezing point of the solution will be:

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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

The correct option is B 1.12 KHX⇌H+X−1001−ααα ∴i=1−α+α+α=1+α=1+0.2=1.2 So, the lowering of freezing point of the solution, ΔTf=i×Kf×m=1.2×1.86×0.5 =1.116 K≈1.12 K

$0.5$ 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