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Relative Lowering of Vapour Pressure

0.2 Molal aci...

Question

0.2 Molal acid $H X$ is $20 %$ ionized in solution, $K_{f} = 1.86 K$ molality $^{- 1}$ . The freezing point of solution (in $^{o} C$ ) is:

- 0.45

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- 0.90

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- 0.31

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- 0.53

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Solution

The correct option is B $- 0.45$
$H X ⟶ H^{+} + X^{-}$
initially $1$ $0$ $0$
At equilibrium $(1 - 0.2) 0.2$ $0.2 \times 0.2$ $0.2 \times 0.2$
$0.16 M$ $0.4 M$ $0.4 M$

Total moles of solution of equilibrium
$= 0.16 + 0.4 + 0.4$
$= 0.24$
${Δ T}_{f} = K_{f} m$
$= 1.86 \times 0.24$
$= 0.45$
So, freezing point of the solution $= 0 - Δ T_{f} = 0 - 0.45 = - 0.45$

Hence, the correct option is $A$

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Similar questions

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

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]

0.5

molal aqueous solution of a weak acid

(H X)

20 %

ionized. If

K_{f}

for the water is

1.86 K k g m o l^{- 1}

, the lowering in freezing point of the solution is:

Q. In a

0.2

molal aqueous solution of a weak acid,

H X

. the degree of ionization is

0.3

. If

K_{f}

for water is

1.85

, find the freezing point of the solution.

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0.2 Molal acid HX is 20% ionized in solution, Kf=1.86K molality−1. The freezing point of solution (in oC) is:

The correct option is B −0.45 HX ⟶ H+ + X−initially 1 0 0At equilibrium (1−0.2)0.2 0.2×0.2 0.2×0.2 0.16M 0.4M 0.4MTotal moles of solution of equilibrium=0.16+0.4+0.4=0.24ΔTf=Kfm =1.86×0.24 =0.45So, freezing point of the solution =0−ΔTf=0−0.45=−0.45Hence, the correct option is A

0.2 Molal acid $H X$ is $20 %$ ionized in solution, $K_{f} = 1.86 K$ molality $^{- 1}$ . The freezing point of solution (in $^{o} C$ ) is: