For a dilute solution containing 2.5g of a non-volatile non-electrolyte solute 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)=(1000K_(b)w_(2))/(w_(1)xxDeltaT_(b))=(1000xx0.76xx2.5)/(100xx2)=9.5`
`(p^(@)-p_(s))/(p^(@))=x_(2)=(n_(2))/(n_(1)+n_(2))=(n_(2))/(n_(1))" "(" as " n_(2) lt lt n_(1))`
`=(w_(2)//M_(2))/(w_(1)//M_(1))`
`(760-p_(s))/(760)=(2.5//9.5)/(100//18)=0.047`
`or 760-p_(s)=760xx0.047=35.7`
`or" "p_(s)=724.3mm`