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$2 \mathrm{~g}$ 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..........
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$0.3 \mathrm{~K}$
$\Delta T_f=K_f \times \frac{w_1 \times 1000}{w_2 \times m_1}$
$\begin{aligned} & w_1=\text { weight of solute } \\ & w_2=\text { weight of solvent } \\ & m_1=\text { molar mass of solute } \\ & K_f=4.3 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\end{aligned}$
Now, $\Delta T_f=4.3 \times \frac{2 \times 1000}{57.3 \times 500}$
$=\frac{17.2}{57.3}=0.30 \mathrm{~K}$
$\begin{aligned} & w_1=\text { weight of solute } \\ & w_2=\text { weight of solvent } \\ & m_1=\text { molar mass of solute } \\ & K_f=4.3 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\end{aligned}$
Now, $\Delta T_f=4.3 \times \frac{2 \times 1000}{57.3 \times 500}$
$=\frac{17.2}{57.3}=0.30 \mathrm{~K}$
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