Question

Elevation in boiling point of an aqueous urea solution is ${0.52}^{o}C$. $(K_2=0.52Kmol-1kg)$
Then, mole fraction of urea in the solution is:

• A. $0.982$
• B. $0.0567$
• C. $0.943$
• D. $0.018$

Answer 1

The correct option is D $0.018$

As we know that, depression in boiling point,

$\Delta T_b=k_b.m$

where,

$\Delta T_b$ = Elevation in boiling point = $0.52$

$k_b$ = molal elevation of boiling point = $0.52$

$m$ = molality of solution = $?$

$\therefore 0.52=0.52\times m$

$\Rightarrow m=1$

As,

$m=\frac{\text{moles of solute}}{\text{Kg of solvent}}=1$

$\Rightarrow$ As the molality is 1, it signifies that 1 mole of urea is present in 1 kg of water.

$\therefore$ no. of moles of water in the solution = $\frac{1000}{18}=55.55;(\because \text{18 is the molar mass of water})$

As we know,

mole fraction of solute or solvent in the solution = $\frac{\text{no. of moles of solute or solvent}}{\text{total no. of moles of solution}}$

$\therefore$ mole fraction of urea in water $=\frac{\text{no. of moles of urea in the solution}}{\text{total no. of moles}}$

$\Rightarrow$ mole fraction of urea in water $=\frac{1}{1+55.55}=\frac{1}{56.55}=0.0176\approx 0.018$

Hence, the mole fraction of urea in the solution is 0.018.

Hence, the correct option is $D$