Question

An aqueous solution of urea has freezing point $-{0.52}^{o}C.$ Then its molality will be:

($K_f={1.86}^{0}C/m$)

• A. 0.28
• B. 2
• C. 0.57
• D. 0.82

Answer 1

Answer: (A)

0.28

$\Delta T_f=K_f\times molality$

Here, $\Delta T_f=T_{solvent}-T_{solution}$

$\Delta T_f=0-(-0.52)={0.52}^{0}C$

$\therefore 0.52=1.86\times molality$ $(\because K_{f}forwater=1.86)$

$\therefore molality=\frac{0.52}{1.86}=0.28m$