Q12 / 90 · JEE Main 2011 · Chemistry
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to $4 \mathrm{~kg}$ of water to prevent it from freezing at $-6^{\circ} \mathrm{C}$ will be :
$\left[\mathrm{K}_{\mathrm{t}}\right.$ for water $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, and molar mass of ethylene glycol $=62 \mathrm{~g} \mathrm{~mol}^{-1}$ )
- A$204.30 \mathrm{~g}$
- B$400.00 \mathrm{~g}$
- C$304.60 \mathrm{~g}$
- Correct answerD$804.32 \mathrm{~g}$
Solution
$$
\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times \mathrm{m}=\mathrm{K}_{\mathrm{f}} \times \frac{\mathrm{w}_2 \times 1000}{\mathrm{w}_1 \times \mathrm{m}_2}
$$
$w_1 \& w_2=$ wt of solvent \& solute respecting $m_2=m w$ of solute
$$
\Delta \mathrm{T}_{\mathrm{f}}=0^{\circ}-\left(-6^{\circ}\right)=6=1.86 \times \frac{\mathrm{w}_2 \times 1000}{4000 \times 62}
$$
Therefore $w_2=800 \mathrm{~g}$
Hence, the correct answer is option: 'D'.