If the vapor pressure of a dilute aqueous solution of glucose is `750 mm` of Hg at `373 K`, then molality of solute is

Options

A. `0.7432 m`
B. `0.8976 m`
C. `0.6753 m`
D. `0.9811 m`

Answer

1

Explanation

According to 2nd Raoult's law
`(P_(1)^(0)-P)/P_(1)^(0)=chi_(2)`
Since solvent is `H_(2)O` and at `373 K` (normal boiling point of water) vapor pressure of pure water is `760 mmHg`, we have
`((760-750) mmHg)/(760 mmHg)`
or `chi_(2)=0.0132`
By definition, `chi_(2)=n_(2)/(n_(1)+n_(2))=n_(2)/n_("total")=0.0132/1`
where `n_(2)` is no. of moles of glucose and `n_(1)` is no. of moles of `H_(2)O`.
Considering 1 mol of solution, we have
`n_(2)=0.0132`
`n_(1)=1-0.0132=0.9868`
since `n_(H_(2)O)=(mass_(H_(2)O))/(molar mass_(H_(2)O))`
`mass_(H_(2)O)=(n_(H_(2)O))(molar mass_(H_(2)O))`
`=(0.9868 mol)(18 g mol^(-1))`
`=17.76 g`
Now, molality `=n_(1)/g_(solv)xx(1000 g)/(kg)`
`=(0.0132 mol)/(17.76 g)xx(1000 g)/(kg)`
`=0.7432 mol Kg^(-1)`