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The mass of a non-volatile solute of molar mass $60 \mathrm{~g} \mathrm{~mol}^{-1}$ that should be dissolved in $126 \mathrm{~g}$ of water to reduce its vapour pressure to $99 \%$ will be
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$4.2 \mathrm{~g}$
Given, molar mass of solute, $\mu_{2}=60 \mathrm{~g} / \mathrm{mol}$
Mass of water $\left(W_{1}\right)=126 \mathrm{~g}^{\circ}$
Molar mass of water $\left(M_{1}\right)=18 \mathrm{~g} / \mathrm{mol}$
$p_{\text {Solution }}\left(p_{s}\right)=0.99 p^{\circ}$
Relative lowering of vapour pressure
$\quad \frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{W_{2}}{M_{2}} \frac{M_{1}}{W_{1}}$
$\quad \frac{p^{\circ}-0.99 p^{\circ}}{W_{2} \times 18}=\frac{0.1 p^{\circ}}{60 \times 126}=\frac{W_{2} \times 18}{p^{\circ}} 60 \times 126$
$\Rightarrow \quad \frac{60 \times 126}{18 \times 100}=4.2 \mathrm{~g}$
Mass of water $\left(W_{1}\right)=126 \mathrm{~g}^{\circ}$
Molar mass of water $\left(M_{1}\right)=18 \mathrm{~g} / \mathrm{mol}$
$p_{\text {Solution }}\left(p_{s}\right)=0.99 p^{\circ}$
Relative lowering of vapour pressure
$\quad \frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{W_{2}}{M_{2}} \frac{M_{1}}{W_{1}}$
$\quad \frac{p^{\circ}-0.99 p^{\circ}}{W_{2} \times 18}=\frac{0.1 p^{\circ}}{60 \times 126}=\frac{W_{2} \times 18}{p^{\circ}} 60 \times 126$
$\Rightarrow \quad \frac{60 \times 126}{18 \times 100}=4.2 \mathrm{~g}$
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