20 g of naphthoic acid (C_{11}H_8O_2) (M_w = | vedclass

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

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$0.5$

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(C) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m$. First,calculate the molality $(m)$ of the solution: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{20 \ g / 172 \ g/mol}{0.050 \ kg} = \frac{0.11628 \ mol}{0.050 \ kg} = 2.3256 \ mol/kg$. Given $\Delta T_f = 2 \ K$ and $K_f = 1.72 \ K \ kg \ mol^{-1}$,substitute these values into the equation: $2 = i 	imes 1.72 	imes 2.3256$. Solving for $i$: $i = \frac{2}{1.72 	imes 2.3256} = \frac{2}{4} = 0.5$. Thus,the Van't Hoff factor is $0.5$.

$20 \ g$ of naphthoic acid $(C_{11}H_8O_2)$ $(M_w = 172 \ g/mol)$ is dissolved in $50 \ g$ of benzene. The freezing point depression of $2 \ K$ is observed for the solution. The Van't Hoff factor $(i)$ will be ($K_f$ for benzene $= 1.72 \ K \ kg \ mol^{-1}$)

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