The freezing point of a 0-025M solution of a weak acid HA is -0-060-C- The percent dissociation is- K b H 2 O - 1-86 K Kg mol -1

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Standard XII

Chemistry

Elevation in Boiling Point

The freezing ...

Question

The freezing point of a $0.025 M$ solution of a weak acid $H A$ is $- {0.060}^{\circ} C$ . The percent dissociation is:
$K_{b} (H_{2} O) = 1.86 K K g {m o l}^{- 1}$

29

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Solution

The correct option is C $29$
Using the relation, $Δ T_{f} = i K_{f} m$

$i = \frac{Δ T_{f}}{K_{f} m}$

$i = \frac{0.060}{1.86 \times 0.025} = 1.29$

The weak acid will dissociate as:

$H X$ $⇌$ $H^{+} + X^{-}$

$0.025 - α$ $α$ $α$

Hence, $i = 1 + α$

$\Rightarrow α = 0.29$

Percentage of dissociation $= 29 %$

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The freezing point of a 0.025M solution of a weak acid HA is −0.060∘C. The percent dissociation is: Kb(H2O)=1.86 K Kgmol−1

The correct option is C 29Using the relation, ΔTf=iKfmi=ΔTfKfmi=0.0601.86×0.025=1.29The weak acid will dissociate as: HX ⇌ H++X−0.025−α α αHence, i=1+α⇒α=0.29Percentage of dissociation =29%

The freezing point of a $0.025 M$ solution of a weak acid $H A$ is $- {0.060}^{\circ} C$ . The percent dissociation is:
$K_{b} (H_{2} O) = 1.86 K K g {m o l}^{- 1}$