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(D) The formula for freezing point depression is $\Delta T_f = i \cdot K_f \cdot m$. Here,$i = 2.67$,$K_f = 1.8 \, K \, kg \, mol^{-1}$. The molality

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$243$

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(D) The formula for freezing point depression is $\Delta T_f = i \cdot K_f \cdot m$. Here,$i = 2.67$,$K_f = 1.8 \, K \, kg \, mol^{-1}$. The molality $m$ is calculated as: $m = \frac{	ext{mass of solute}}{	ext{molar mass of solute}} 	imes \frac{1000}{	ext{mass of solvent in grams}}$. For $38\%$ by weight $H_2SO_4$,mass of $H_2SO_4 = 38 \, g$,mass of water = $100 - 38 = 62 \, g$. $m = \frac{38}{98} 	imes \frac{1000}{62} \approx 6.254 \, mol \, kg^{-1}$. $\Delta T_f = 2.67 	imes 1.8 	imes 6.254 \approx 30.05 \, K$. Freezing point of pure water = $273.15 \, K$. Freezing point of solution = $273.15 - 30.05 = 243.1 \, K$. Rounding to the nearest integer,we get $243 \, K$.

Lead storage battery contains $38\%$ by weight solution of $H_2SO_4$. The van't Hoff factor is $2.67$ at this concentration. The temperature in Kelvin at which the solution in the battery will freeze is $..........$ (Nearest integer). Given $K_f = 1.8 \, K \, kg \, mol^{-1}$.

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