Question

If 12 g of urea is dissolved in 500 g water then the freezing point of the solution will be:
(K_f of water = 1.86 K kg mol^–1)

• A. –1.74°C
• B. –0.22°C
• C. –0.74°C
• D. –1.21°C

Answer 1

The correct option is C –0.74°C

$m=\frac{12\times 1000}{60\times 500}$

$\Delta T_f=K_f\times m$

$=1.86\times \frac{12\times 1000}{60\times 500}=0.744$ (For urea, i = 1)

$T_f=-{0.74}^{\circ }C$

Freezing point of solution $=-{0.74}^{\circ }C$