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On dissolving 0.5 g of non-volatile, non...

On dissolving 0.5 g of non-volatile, non-ionic solute to 39 g of benzene, its vapour pressure decreases from 650 mm of Hg to 640 mm of Hg. The depression of freezing point of benzene (in K) upon addition of the solute is __________.
(Given data: Molar mass & molar freezing point depression is 78 g `mol^(-1) & 5.12 K kg mol^(-1)`]

Text Solution

Verified by Experts

The correct Answer is:
1.03
From Raoult.s law, the relative lowering in vapour pressure is given by
`(p_(B)^(o)-p_(s))/(p_(s))=(n_(A))/(n_(A)+n_(B)) rArr (650-640)/(650)=(n_(A))/(n_(A)+0.5) ("as n"_(B)=(39)/(78)=0.5"mol")`
So, we obtain `n_(A)=5/(640)`. The colligative property, depression in freezing point is given by
`triangleT_(f)=iK_(f)m=1 xx 5.12 xx (5 xx 1000)/(640 xx 39)=1.024 K`
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