A very small amount of a non volatile solute(that does not dissociate) is dissolved in $56.8$ cc of benzene (density $0.889$ g/cc). At room temperature the vapour pressure of this solution is $98.88$ mm Hg while that of benzene is $100$ mm Hg. Find the molality of this solution freezing temperature of this solution is $0.73$ degree lower than that of benzene, hat is the value of molal freezing point depression constant of benzene.

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Solution

$\frac{P^{0} - P}{P^{0}} = \frac{w}{m} \times \frac{M}{W}$
$Δ T_{f} = K_{f} \times m$
put the given values
$\frac{1000 - 98.88}{98.88} = \frac{w \times 78 \times 1000}{m \times W \times 1000}$
$\frac{w \times 100}{m \times W} = \frac{1.12 \times 1000}{78 \times 98.8}$
$Δ T_{f} = K_{f} \times m$
$0.73 = K_{f} \times 0.1452$
$K_{f} = 5.027 K m o l a l i t y^{- 1}$ ( $\frac{w}{m} \times \frac{1000}{W} = M o l a l i t y$ )

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P0−PP0=wm×MWΔTf=Kf×mput the given values 1000−98.8898.88=w×78×1000m×W×1000w×100m×W=1.12×100078×98.8ΔTf=Kf×m0.73=Kf×0.1452Kf=5.027Kmolality−1 (wm×1000W=Molality)