A $0.2 m o l a l$ aqueous solution of a weak acid $H X$ is $20 %$ ionized. The freezing point of the solutions is :
[ $k_{f} = 1.86 k k g m o l e^{- 1}$ for water]

- {0.45}^{\circ} C

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- {0.9}^{\circ} C

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- {0.31}^{\circ} C

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- {0.53}^{\circ} C

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Solution

The correct option is A $- {0.45}^{\circ} C$
We know, $Δ T_{f} = i K_{f} m$

$∴ Δ T_{f} = 1.86 \times 0.2 \times 1.2$

$Δ T_{f} = 0.45 45^{\circ} C$

Freezing point of the solution $= 0 - 0.45 = - 0. 45^{\circ} C$

Therefore, the correct option is A.

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

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

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A 0.2molal aqueous solution of a weak acid HX is 20% ionized. The freezing point of the solutions is :[kf=1.86kkgmole−1 for water]

The correct option is A −0.45∘CWe know, ΔTf=iKfm∴ΔTf=1.86×0.2×1.2ΔTf=0.4545∘CFreezing point of the solution =0−0.45=−0.45∘CTherefore, the correct option is A.

A $0.2 m o l a l$ aqueous solution of a weak acid $H X$ is $20 %$ ionized. The freezing point of the solutions is :
[ $k_{f} = 1.86 k k g m o l e^{- 1}$ for water]

The correct option is A $- {0.45}^{\circ} C$
We know, $Δ T_{f} = i K_{f} m$

$∴ Δ T_{f} = 1.86 \times 0.2 \times 1.2$

$Δ T_{f} = 0.45 45^{\circ} C$

Freezing point of the solution $= 0 - 0.45 = - 0. 45^{\circ} C$

Therefore, the correct option is A.