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Question

A solution containing 0.1 mole of naphthalene and 0.9 mole of benzene is cooled out until some benzene freezes out. The solution is then decanted off from the solid and warmed up to 353 K where its vapour pressure was found to be 670 torr. The freezing point and boiling point of benzene are 278.5 K and 353 K respectively and its enthalpy of fusion is 10.67 kJ mol1. Calculate the depression in freezing point for original solution to the nearest integer. Assume ideal behaviour

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Solution

P0=760mm at boiling point 353 K and P0PSPS=w×Mm×W
760670670=wN×MBmN×WB=0.1×78WB
WB=58.06g
Also in original solution WB=0.9×78=70.2g
C6H6 frozen out =70.258.06=12.14g
Now, ΔT (for original solution)=Kf×molality
=RT2b1000l×molality(molality=0.158.06×1000)
=8.314×(278.5)21000×136.8×1000×0.158.06(1=10.67×10378=136.8J/g)
=8.12K

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