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

Options

A. 724
B. 740
C. 736
D. 718

Answer

A

Explanation

`M_(B) = (k_(b) xx 1000 xx w_(b))/( Delta T_(b) xx w_(A))`
` = (0.76 xx 1000 xx 2.5)/( 2 xx 100) = 9.5`
` (p_(@)^(A) - p_(A))/(p_(A)^(@)) = x_(2) (n_(2))/(n_(1)) (as" "n_(2) lt lt n_(1))`
`(p_(A)^(@) - p_(A))/(p_(A)^(@)) = (w_(1)//M_(2))/(w_(1)//M_(1))`
`(760 - p_(A))/(760) = (2.5//9.5)/(100 //18) = 0.0474`
`760-p_(A) = 0.0474 xx 760 = 36`
` or P_(A) = 760 - 36 = 724` mm.