Question

The boiling point of pure water is $100{}^{\circ }$C. Calculate the boiling point of an aqueous solution containing $0.6\text{g}$ of urea (molar mass = $60{\text{g mol}}^{-1}$) in $100\text{g}$ of water. $K_b$ for water is $0.52{\text{K kg mol}}^{-1}$.

• A. $100.052{}^{\circ }\text{C}$
• B. $200.055{}^{\circ }\text{C}$
• C. $373.052{}^{\circ }\text{C}$
• D. $273.052{}^{\circ }\text{C}$

Answer 1

The correct option is A $100.052{}^{\circ }\text{C}$

Since the molality of the solute is given as follows,
$\text{Molality (m)}=\frac{\text{Number of moles of the solute}\times 1000}{\text{Weight of the solvent}}$

$\Rightarrow \text{Molality (m)}=\frac{\frac{\text{given mass of the solute}}{\text{molar mass of the solute}}\times 1000}{\text{Weight of the solvent}}$

$\Rightarrow \text{Molality}=\frac{\frac{0.6}{60}\times 1000}{100}$

$\Rightarrow \text{Molality}=0.1\text{m}$

The elevation of the boiling point is given by $\Delta T_b=K_{b}m$
$\Rightarrow \Delta T_b=0.52\times 0.1$
$\Rightarrow \Delta T_b=0.052\text{K}$

$\Rightarrow \Delta T_b=T_b-T_b^{\circ }$
$\Rightarrow 0.052=T_b-100$
$T_b=100.052{}^{\circ }\text{C}$