Question

$31\, g$ of ethylene glycol $\left( C _{2} H _{6} O _{2}\right)$ is mixed with $500\, g$ of solvent $\left(K_{f}\right.$ of the solvent is $2\, K\,kg\, mol ^{-1}$ ). What is the freezing point of the solution (in $K$ )? (Freezing point of solvent $=273\, K )$

• A. 272
• B. 271
• C. 270
• D. 274
• E. 275

Answer 1

Answer: (B)

Given, $W_{B}=$ mass of solute (ethylene glycol) $=31\, g$

$W_{A}=$ mass of solvent $=500\, g$

$M_{B}=$ Ethylene glycol $\left( C _{2} H _{6} O _{2}\right)$

$=(2 \times 12+6+2 \times 16)=62\, g$

$K_{f}=2\,K\,kg\, mol ^{-1}$

$\Delta T_{f}=K_{f} \frac{W_{B}}{M_{B}} \times \frac{1000}{W_{A}}$

$\Delta T_{f}=\frac{2 \times 31 \times 1000}{62 \times 500}=2\, K$

$\because \Delta T_{f}=T_{f}^{\circ}-T_{f}$

$T_{f}=T_{f}^{\circ}-2$

$(\because$ Depression in freezing point $=2\, K )$

$\therefore T_{f}=273-2=271\, K$