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What is freezing point of a liquid ? The freezing point of pure benzene is 278.4 K. Calculate the freezing point of the solution when 2.0 gram of a solute having molecular weight 100 gram is added to 100 gram of benzene. (K_(f) for benzene =5.12 kg "mol"^(-1))

Answer»

Solution :FREEZING POINT of aliquid is a temperature at whichthe vapour pressure of solid is EQUAL to the vapour pressure of the LIQUID.
Given,
`(T^(@))` Freezing point of benzene = 278.4 K
(T) Freezing pointof solution = ?
`(W_(2))` Mass of solute `=2.0 g= 2.0xx10^(-3)kg`
`(M_(2))` Molar mass of solute `=100g = 0.1 kg `
`K_(f)` of benzene `=5.12` kg `"mol"^(-1)`
`(W_(1))` Mass of solvent `=100 g = 0.1 kg`
`Delta T_(f)=(W_(2)xxK_(f)xx1000)/(M_(2)xxW_(1))`
`T^(@)-T=(2.0xx5.12xx10^(-3))/(0.100xx0.1)`
`278.4-T=1.024`
`278.4-T=1.024K`
`T=-1.024+278.4`
`=277.376` K
Freezing point of solution is 277.376 K.


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