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 constant of benzene is "5.12 K kg mol"^(-1). Find the molar mass of solute.

Answer»

SOLUTION :Here, we are given `w_(2)=1.00 g, w_(1)=50, DeltaT_(f)=0.40 K, K_(f)="5.12 K kg MOL"^(-1)`
Substituting these values in the formula `M_(2)=(1000K_(f)w_(2))/(w_(1)DeltaT_(f)),` we get
`M_(2)=("1000 g kg"^(-1)xx"5.12 kg mol"^(-1)xx1.0g)/(50 g xx0.40 K)="256 g mol"^(-1)`


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