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)`.

Explanation

Here, we are given `" " w_(2)=1.80 g, w_(1)=90 g, " " Delta T_(b)=354.11-353.23 K=0.88 K`
`" " K_(b)=2.53 " K kg mol"^(-1)`
Substituting these values in the formula,
`M_(2)=(1000 K_(b)w_(2))/(w_(1)Delta T_(b))`, we get `M_(2)=(1000 " g kg"^(-1)xx2.53" K kg mol"^(-1)xx1.80 g)/(90g xx 0.88 K)=58 "g mol"^(-1)`