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The weight in grams of a non-volatile solute (mol. wt. 60) to be dissolved in $90 \mathrm{~g}$ of water to produce a relative lowering of vapour pressure of 0.02 is
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$6$
Relative lowering of vapour pressure,
$\begin{aligned} \frac{p^{\circ}-p_s}{p^{\circ}} & =x_A \\ & =\frac{\frac{w_A}{m_A}}{\frac{w_A}{m_A}+\frac{w_B}{m_B}}\end{aligned}$
(where, $w_A$ and $m_A$ are the mass and molar mass of solute and $w_B$ and $m_B$ are the mass and molar mass of water.)
$\therefore \quad \begin{aligned} 0.02 & =\frac{\frac{x}{60}}{\frac{x}{60}+\frac{90}{18}} \\ 0.02 & =\frac{\frac{x}{60}}{\frac{x}{60}+5} \\ \frac{1}{0.02} & =\frac{\frac{x}{60}+5}{\frac{x}{60}} \\ 50 & =1+\frac{5 \times 60}{x} \\ 49 & =\frac{300}{x} \\ x & =\frac{300}{49}=6 \mathrm{~g}\end{aligned}$
$\begin{aligned} \frac{p^{\circ}-p_s}{p^{\circ}} & =x_A \\ & =\frac{\frac{w_A}{m_A}}{\frac{w_A}{m_A}+\frac{w_B}{m_B}}\end{aligned}$
(where, $w_A$ and $m_A$ are the mass and molar mass of solute and $w_B$ and $m_B$ are the mass and molar mass of water.)
$\therefore \quad \begin{aligned} 0.02 & =\frac{\frac{x}{60}}{\frac{x}{60}+\frac{90}{18}} \\ 0.02 & =\frac{\frac{x}{60}}{\frac{x}{60}+5} \\ \frac{1}{0.02} & =\frac{\frac{x}{60}+5}{\frac{x}{60}} \\ 50 & =1+\frac{5 \times 60}{x} \\ 49 & =\frac{300}{x} \\ x & =\frac{300}{49}=6 \mathrm{~g}\end{aligned}$
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