Question

Relative lowering of vapour pressure of a non-volatile solute $X$ in water is $0.15 .$ If $K _{ b }\left( H _{2} O \right)=0.52^{\circ} mol ^{-1}$ $kg ^{-1}$
$\left(\Delta T _{ b }\right)=$ elevation in boiling point assuming a dilute solution
$\left(\Delta T _{ b }\right)'=$ elevation in boiling point assuming a solution of moderate concentration then, $\left(\Delta T _{ b }\right)-\left(\Delta T _{ b }\right)'=$

• A. Zero
• B. $-0.87^{\circ}$
• C. $-0.08^{\circ}$
• D. $0.16^{\circ}$

Answer 1

Answer: (C)

By Raoult's law,

$\frac{\Delta p}{p^{\circ}}=0.15=\chi_{\text {solute }}$

$\Delta T_{b}=\frac{100 K_{b} w_{1}}{m_{1} w_{2}}$

If solution is dilute,

$\chi_{\text {solute }}=\frac{n_{1}}{n_{1}+n_{2}}=\frac{n_{1}}{n_{2}}=\frac{w_{1} m_{2}}{m_{1} w_{2}}$

$\therefore \frac{w_{1} m_{2}}{m_{1} w_{2}}=0.15 $

$\therefore \frac{w_{1}}{m_{1} w_{2}}=\frac{0.15}{18}$

$\therefore \Delta T_{b}=\frac{1000 \times 0.52 \times 0.15}{18}=4.33^{\circ}$

If solution is moderate concentration,

then $\frac{\Delta p}{p^{\circ}}=\frac{n_{1}}{n_{1}+n_{2}}=\frac{0.15}{1}$

Thus, $ n_{1} =0.15 \,mol $

$n_{2} =0.85 \,mol =0.85 \times 18 \,g $

$=\frac{0.85 \times 18}{1000}=\frac{15.3}{1000} kg =0.0153\, kg$

$\therefore $ Molality $=\frac{0.15\, mol }{0.015 \,kg }=10 \,mol \,kg ^{-1}$

$\left(\Delta T_{b}\right)'=K_{b} \times$ molality

$=0.52 \times 10=5.2^{\circ}$

$\therefore \left(\Delta T_{b}\right)-\left(\Delta T_{b}\right)'$

$=4.33-5.2=-0.87^{\circ}$