When 1.0 g of KCl is dissolved in 200 g wat | vedclass

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(C) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. First,calculate the molality $(m)$ of the solution: Moles of

Vedclass · Accepted Answer

$2.0$

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(C) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. First,calculate the molality $(m)$ of the solution: Moles of $KCl = \frac{	ext{mass}}{	ext{molar mass}} = \frac{1.0 \ g}{74.5 \ g \ mol^{-1}} \approx 0.0134 \ mol$. Mass of solvent in $kg = 200 \ g = 0.2 \ kg$. Molality $(m) = \frac{0.0134 \ mol}{0.2 \ kg} = 0.067 \ mol \ kg^{-1}$. Now,substitute the values into the depression in freezing point equation: $0.24 \ K = i 	imes 1.86 \ K \ kg \ mol^{-1} 	imes 0.067 \ mol \ kg^{-1}$. $i = \frac{0.24}{1.86 	imes 0.067} \approx \frac{0.24}{0.1246} \approx 1.926$. Rounding to the nearest option,the Van't Hoff factor $(i)$ is approximately $2.0$.

When $1.0 \ g$ of $KCl$ is dissolved in $200 \ g$ water,the decrease in freezing point of such solution is $0.24 \ K$,calculate Van't Hoff factor $(i)$ for such solution. $K_f$ of water $=$ $1.86 \ K \ kg \ mol^{-1}$.

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