Question

Relative lowering of vapour pressure of a non-volatile solute $X$ in water is $\mathrm{0.15.}$ If $K_b\left(H_{2}O\right)=0.52^{\circ }mo{l}^{-1}$ $k{g}^{-1}$
$\left(\Delta T_b\right)=$ elevation in boiling point assuming a dilute solution
${\left(\Delta T_b\right)}^{\prime }=$ elevation in boiling point assuming a solution of moderate concentration then, $\left(\Delta T_b\right)-{\left(\Delta T_b\right)}^{\prime }=$

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

Answer 1

Answer: (C)

By Raoult's law,

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

$\Delta T_b=\frac{100K_b{w}_1}{m_1{w}_2}$

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, $n_1=0.15mol$

$n_2=0.85mol=0.85\times 18g$

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

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

${\left(\Delta T_b\right)}^{\prime }=K_b\times$ molality

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

$\therefore \left(\Delta T_b\right)-{\left(\Delta T_b\right)}^{\prime }$

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