Question

Relative lowering of vapour pressure of a dilute aqueous solution of glucose is found to be $0.018$. Hence, elevation in boiling point is:
(Given, that $1$ molal aqueous urea solution boils at $00.54{}^{o}C$ at $1$ atm pressure)

• A. $0.018{}^{o}C$
• B. $0.18{}^{o}C$
• C. $0.54{}^{o}C$
• D. $0.03{}^{o}C$

Answer 1

Answer: (C)

Given,
$\Delta T_b\text{for 1 m urea solution}=0.54{}^{\circ }C$
$\Rightarrow 0.54{}^{\circ }C=K_b\times \text{molality}$
$\Rightarrow K_b=0.54{}^{\circ }C{\text{kg mol}}^{–1}$
Again, relative lowering of vapour pressure,
$\frac{\Delta p}{p^o}=\frac{w_2{m}_1}{m_2{w}_1}$
Where, $w_2=\text{mass of solute}$
$w_1=\text{mass of solvent}$
$m_1=\text{Molar mass of solute}$
$m_2=\text{Molar mass of solvent}$
$\therefore \frac{w_2}{m_2{w}_1}=\frac{\Delta p}{p^o{m}_1}=\frac{0.018}{18}$

$\Delta T_b\left(\text{glucose}\right)=K_{b}m=\frac{1000K_b{w}_2}{m_2{w}_1}=\frac{1000\times 0.54\times 0.018}{18}=0.54{}^{\circ }C$