1.

1.00 g of a non - electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constnat of benzene is 5.12 K kg mol^(-1). Find the molar mass of the solute.

Answer»

Solution :SUBSTITUTING the VALUES of various terms INVOLVED in following EQUATION, we get
`M_(2)=(K_(f)xx w_(2)xx 1000)/(Delta T_(f)xx w_(1))`
`M_(2)=(5.12"K kg mol"^(-1)xx1.00g xx 1000 g kg^(-1))/(0.40xx50 g)`
`= 256g mol^(-1)`
THUS, molar mass of the solute `= 256 g mol^(-1)`


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