If solution containing `0.15 g` of solute dissolved in `15 g` of solvent boils at a temperature higher by `0.216^(@)C` than that of pure solvent, the molecular mass of the substance is `(K_(b)=2.16^(@)C)`

Options

A. `10 u`
B. `1.01 u`
C. `10.1 u`
D. `100 u`

Answer

4

Explanation

Assuming the solute to be a non-electrolyte, we have
`T_(b)=K_(b)m=(K_(b) w_(2))/(M_(2)w_(1))xx(1000 g)/(kg)`
`0.216^(@)C=((2.16^(@)C kg mol^(-1))(0.15 g)(1000 g kg^(-1)))/(M_(2)(15 g))`
`M_(2)=((2.16)(0.15)(1000))/((15)(0.216)) g mol^(-1)`
`=100 g mol^(-1)`
Thus, molecular mass of solute is `100 u`.