Question

The boiling point $T_b^{\ast }$ of a solvent becomes $T_b$ on addition of $X_1$ mole fraction of solute. Heat of vaporisation of solvent is $\Delta H_{vap}$. The relation between elevation in b.pt. $\Delta T_b(=T_b-T_B^{\ast })$ can be given by:
$\frac{\Delta T_b}{T_b\times T_b^{\ast }}=-\frac{RIn{X}_1}{\Delta H_{vap}}$
A graph plotted between $lo{g}_{10}{X}_1$vs.$\frac{1}{T_b}$ gives :

• A. a straight line with slope $\frac{R}{\Delta H_{vap}}$ and intercept $T_b^{\ast }$
• B. a straight line with slope $\frac{2.303R}{\Delta H_{vap}}$ and intercept $\frac{1}{T_b}$
• C. a straight line with slope $\frac{\Delta H_{vap}}{2.303R}$ and intercept $\frac{1}{T_b^{\ast }\times 2.303}$
• D. a straight line with slope $\frac{\Delta H_{vap}}{2.303R}$ and intercept $-\frac{\Delta H_{vap}}{2.303R{T}_b^{\ast }}$

Answer 1

The correct option is D a straight line with slope $\frac{\Delta H_{vap}}{2.303R}$ and intercept $-\frac{\Delta H_{vap}}{2.303R{T}_b^{\ast }}$

A graph plotted between $lo{g}_{10}{X}_{1}vs.\frac{1}{T_b}$ gives a straight line with slope $\frac{\Delta H_{vap}}{2.303R}$ and intercept $-\frac{\Delta H_{vap}}{2.303R{T}_b^{\ast }}$.

$\frac{T_b-T_b^{\ast }}{T_b\times T_b^{\ast }}=-\frac{2.303Rlo{g}_{10}{X}_1}{\Delta H_{vap}}$
$\frac{1}{T_b^{\ast }}-\frac{1}{T_b}=-\frac{2.303Rlo{g}_{10}{X}_1}{\Delta H_{vap}}$
or
$\frac{1}{T_b}-\frac{1}{T_b^{\ast }}=\frac{2.303Rlo{g}_{10}{X}_1}{\Delta H_{vap}}$
$lo{g}_{10}{X}_1=\frac{\Delta H_{vap}}{2.303R{T}_b}-\frac{\Delta H_{vap}}{2.303R{T}_b^{\ast }}$