The boiling point of benzene is 353.23 K. 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)`.

Explanation

The elevation `(Å T_(b))` in the boiling point = `354.11 K – 353. 23 K = 0.88` K Substituting these values in expression (2.33) we get
`M_(2) = (2.53 " K kg mol"^(-1) xx 1.8 g xx 1000 " g kg"^(-1))/(0.88 K xx 90 g ) = 58 " g mol"^(-1)`
Therefore, molar mass of the solute `M_(2) = 58 " g mol"^(-1)`