$2 \mathrm{~g}$ of a non-electrolyte solute (molar mass is $500 \mathrm{~g} \mathrm{~mol}^{-1}$ ) was dissolved in $57.3 \mathrm{~g}$ of xylene. If the freezing point depression constant $K_f$ of xylene is $4.3 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. Then, the depression in freezing point of…

Question

$2 \mathrm{~g}$ of a non-electrolyte solute (molar mass is $500 \mathrm{~g} \mathrm{~mol}^{-1}$ ) was dissolved in $57.3 \mathrm{~g}$ of xylene. If the freezing point depression constant $K_f$ of xylene is $4.3 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. Then, the depression in freezing point of xylene is..........

Aicharya · Accepted Answer

$\Delta T_f=K_f 	imes \frac{w_1 	imes 1000}{w_2 	imes m_1}$ $\begin{aligned} & w_1=	ext { weight of solute } \ & w_2=	ext { weight of solvent } \ & m_1=	ext { molar mass of solute } \ & K_f=4.3 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\end{aligned}$ Now, $\Delta T_f=4.3 	imes \frac{2 	imes 1000}{57.3 	imes 500}$ $=\frac{17.2}{57.3}=0.30 \mathrm{~K}$

$2 \mathrm{~g}$ of a non-electrolyte solute (molar mass is $500 \mathrm{~g} \mathrm{~mol}^{-1}$ ) was…

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