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  4. Calculate the molal depression constant of a solvent which has freezing point 16.6° C and latent heat of fusion 180.75 Jg -1

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Q. Calculate the molal depression constant of a solvent which has freezing point $16.6^{\circ} C$ and latent heat of fusion $180.75$ $Jg ^{-1}$

Solutions

Solution:

$K_{f}=\frac{R T_{f}^{2}}{1000 \times L_{f}}$,

$ R=8.314 \,JK ^{-1} \,mol ^{-1}$

$T_{f}=273+16.6=289.6 \,K$ ;

$ L_{f}=180.75 \,Jg ^{-1}$

$\therefore K_{f}=\frac{8.314 \times 289.6 \times 289.6}{1000 \times 180.75}=3.86$