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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)`) |
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Answer» Freezing point of a liquid is a temperature at which the vapour pressure of solid is equal to the vapour pressure of the liquid. Given, `(T^(@))` Freezing point of benzene = 278.4 K (T) Freezing point of solution = ? `(W_(2))` Mass of solute `=2.0 g= 2.0xx10^(-3)kg` `(M_(2))` Molar mass of solute `=100 g = 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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