The molal lowering of vapor pressure for `H_(2)O` at `100^(@)C` is

Options

A. `20.67 mmHg`
B. `13.45 mmHg`
C. `9.78 mmHg`
D. `30.02 mmHg`

Answer

2

Explanation

Molal lowering of vapor pressure for `H_(2)O` is equal to lowering of vapor pressure of water when 1 mole of any nonvolatile solute is dissolved in `1000 g (1 Kg)` of `H_(2)O` (solvent). Moreover, since `100^(@)C` is the normal `(P_(1)^(@))` is `760 mmHg`.
According to `2^(nd)` Raoult's law, we have
`(P_(1)^(@)-P)/P_(1)^(@)=(DeltaP)/P_(1)^(@)=chi_(2)`
or `DeltaP=P_(1)^(@)(n_(2)/(n_(1)+n_(2)))`
`=(760 mmHg)(1/(1+ 1000 g//18 g mol^(-1)))`
`=(760 mmHg)(1/56.5)`
`=13.45 mmHg`