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Standard XII

Chemistry

Relative Lowering of Vapour Pressure

Total molalit...

Question

Total molality after the addition of $H g (C N)_{2}$ into aqueous solution of $K C N$ .

(0.2842 - 0.095 m)

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(0.4734 - 0.095 m)

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0.2842

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0.4734

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Solution

The correct option is B $(0.4734 - 0.095 m)$
The chemical reaction:
$H g (C N)_{2} + m C N^{-} \to H g (C N)_{m + 2}^{m^{-}}$
Initial moles $0.095$ $0.1892$ $0$
Final moles $0$ $0.1892 - 0.095$ $0.095$
$∴$ Total molality after addition of $H g (C N)_{2}$
$= molality of K^{+} + molality of C N^{-} + molality of H g (C N)_{m + 2}^{m^{-}}$
$= 0.1892 + (0.1892 - 0.0095 m) + 0.095$
$= 0.4734 - 0.095 m$

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