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. 710
C. 736
D. 718

Answer

A

Explanation

`DeltaT_(b)=K_(b).m`
or,`2=0.76xx(2.5xx1000)/(Mxx100)impliesM=(0.76xx25)/(2)`
Now, `(P^(@)-P)/(P^(@))=(n_("solute"))/(n_("solvent"))(n_("solvent")gtgtn_("solute"))`
or, `(760-P)/(760)(((2.5)/(M)))/(((100)/(18)))impliesP=742" mm Hg"`