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 at `100^(@)C` is (take `K_(b)= 0.76 kg mol^(-1)`)

Options

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

Answer

A

Explanation

At `100^(@)C P_(A)^(@)= 760`
`(P_(A)^(@)- P_(S))/(P_(A)^(@))= x_(B)` from elevation in boiling point `Delta T_(b)= k_(f)m`
molality `m= (Delta T_(B))/(k_(f)) = (2)/(0.76)` since the solution is dilute mole fraction `x_(B)= (2)/(76)/(1000)/(18) = (2 xx 18)/(76 xx 1000)`
Hence `(760-P_(S))/(760) = (2 xx 18)/(76 xx 1000), P_(S)= 724 mm Hg`