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_(2) = (K_(b) xx W_(2))/(W_(1) xx Delta T_(b)) xx 1000 = (1000 xx 0.76 xx 2.5)/(100 xx 2) =9.5`
`(p_(1)^(o)-p)/p_(1)^(o) = (W_(2)//M_(2))/(W_(1)//M_(1))`
`:. (760-p)/760 = (2.5 // 9.5)/(100//18)=0.047`
or 760 - p = `760 xx 0.047` = 35.7
or p=724.3 mm