Question

2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at \(25 ^\circ C\). The solution showed a boiling point elevation by \(2 ^\circ C\). Assuming the solute concentration in negligible with respect to the solvent concentration, the vapour pressure of the resulting aqueous solution is _____ mm of Hg (nearest integer) 

[Given : Molal boiling point elevation constant of water \((K_b)\) = 0.52 K. kg \(mol ^{-1}\) ,

1 atm pressure = 760 mm of Hg, molar mass of water = 18 g \(mol^{-1}\)]

Answer 1

Correct answer is : 707

\(2=0.52 \times \mathrm{m}\)

\(\mathrm{m}=\frac{2}{0.52}\)

According to question, solution is much diluted

\(\text { so } \frac{\Delta \mathrm{P}}{\mathrm{P}^{\mathrm{o}}}=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solvent }}} \)

\(\frac{\Delta \mathrm{P}}{\mathrm{P}^{\mathrm{o}}} \) \(=\frac{\mathrm{m}}{1000} \times \mathrm{M}_{\text {solvent }}\)

\(\Delta P = P ^\circ \times \frac {m}{1000} \times M _ {solvent}\)

\(=760 \times \frac{\frac{2}{0.52}}{1000} \times 18=52.615\)

\(P_5 = 760 - 52.615 = 707.385\) mm of Hg