$\mathrm{K}_{\mathrm{f}}$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. If your…

$\mathrm{K}_{\mathrm{f}}$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. If your automobile radiator holds $1.0 \mathrm{~kg}$ of water, how many grams of ethylene glycol $\left(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}_2\right)$ must you add to get the freezing point of the solution lowered to $-2.8^{\circ} \mathrm{C}$ ?
  1. $72 \mathrm{~g}$
  2. $93 \mathrm{~g}$
  3. $39 \mathrm{~g}$
  4. $27 \mathrm{~g}$

Solution

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \cdot \mathrm{m}$ $2.8=1.86 \times \frac{\mathrm{wt}}{62} \times \frac{1000}{1000}$ $\mathrm{Wt}=93 \mathrm{~g}$

Asked in: JEE Main 2012 (Offline)

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