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

Therefore, the freezing point is `-2.3 xx 10^(-2@)C`.
`DeltaT_(b)=2^(@)C`,`W_(2)=2.5 g`
`W_(1)=100 g`,`K_(b)=0.76 K. kg mol^(-1)`
`p_(s)=?`
`DeltaT_(b)=K_(b) xx m`
`2=0.76 xx m rArr m=2/0.76`
We know,
`m=(n_(2) xx 1000)/(n_(1) xx Mw_(1)) (Mw_(1)(H_(2)O)=18)`
`n_(2)/n_(1)=x_(2)=(m xx Mw_(1))/1000=(2 xx 18)/(0.76 xx 1000) =36/760`
`:. (p^(@)-p_(S))/p_(S)=x_(2)=(2 xx 18)/(0.76 xx 1000) =36/760`
`760 - p_(S)=36/760 p_(s)`
`(36/760 p_(s) + p_(s))=760`
`p_(s)(36/760) +1=760`
`1.047 p_(s)=760`
`p_(s)=(760/1.047)`
=`725.6 "torr"~~724 "torr"`