For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g water, the elevation of boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, determine the vapour pressure (mm Hg) of the solution. [Given: `k_b` for water = 0.76K-kg-`mol^(- 1)` ]

Explanation

According to Rault's law.
`/_\T_b=2K(=-2^(@)C),k_b=0.76K*kg*mol^(-1)`
`:." "2=0.76xxm`
`:." "m=2.63mol*kg^(-1)`
`:.` Molar mass of solute M`=(2.5)/(m)xx(1000)/(100)`
`=(2.5)/(2.63)xx(1000)/(100)=9.5`
According to Rault's law, `(p^0-p)/(p^0)=x_2=(n_2)/(n_1)=(W_2M_1)/(W_1M_2)`
or `(1-p)/1=(2.5xx18)/(100xx9.5)`
or, p`=0.953atm=724.28mm" "Hg`
[`:'` 1atm=760mm Hg]