Question

The amount of ice that will separate out on cooling a solution containing $45g$ of ethylene glycol in $250g$ water to $-8.3^{\circ }C$ is $(K_f$ of $H_{2}O=1.86^{\circ }mo{l}^{-1}kg)$

• A. 38.7 g
• B. 161.3 g
• C. 87.4 g
• D. 152.7 g

Answer 1

Answer: (C)

$w_1=$ solute $=50g$ ethylene glycol

$\frac{w_1}{m_1}=\frac{45}{62}mol$

Solvent required to maintain this amount $=w_{2}g$

$\Delta T_f=8.3^{\circ }$

$\therefore \Delta T_f=\frac{1000w_1{K}_f}{m_1{w}_2}$

$\therefore w_2=\frac{1000w_1{K}_f}{m_1\Delta T_f}$

$=\frac{1000\times 45\times 1.86}{62\times 8.3}=162.6g$

Actual $H_{2}O$ taken $=250g$

Thus, ice formed $=250-162.6=87.4g$