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Q.

A very small amount of a non-volatile solute (that does not dissociate) is dissolved in 56.8cm3 of benzene (density  0.889 gcm3). At room temperature, vapour pressure of this solution is 98.88mm Hg while that of benzene is 100mm Hg. If the freezing temperature of this solution is 0.73 degree lower than that of benzene, what is the value of molal freezing point depression constant of benzene?

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answer is 5.

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Detailed Solution

P0PsPs=wm×MW  ,                  10098.8898.88=w×78×1000m×W×1000
molality  (w×1000m×W)=1.12×100078×98.88=0.1452
Also, ΔT=K'f×  molality
0.73 =  K'f×0.1452               K'f=5.028Kmolality1      

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