Question

Elevation in boiling point of an aqueous urea solution is $0.52{}^{\circ }C[K_b=0.52{}^{\circ }mo{l}^{-1}kg].$ Hence, mole fraction of urea in this solution is

• A. 0.982
• B. 0.567
• C. 0.943
• D. 0.018

Answer 1

Answer: (D)

$\because$ $\Delta T_b=molality\times K_b$
$\because$ $0.52=m\times 0.52$
or molality $=1molk{g}^{-1}$
$\therefore$ No. of moles of urea = 1 mol No. of moles of water $<br/>=\frac{1000}{18}=55.55mol$
Mole fraction of urea $=\frac{molesofurea}{molesofurea+molesofwater}=0.018$
$=\frac{1}{1+55.55}$
$=\frac{1}{56.55}$