1.

The boiling a point of benzene is 353.23K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. `K_(b)` for benzene is 2.53 K kg `mol^(-1)`.

Answer» The elevation `(Deltat_(b))` in the boiling point `=354.11K-353.23K=0.88K`
Substituting these values in expression (2.33) we get
`M_(2)=(2.53" K kg "mol^(-1)xx1.8" g "xx1000" g "kg^(-1))/(0.88Kxx90g)=58" g "mol^(-1)`


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