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The vapour pressure in $\mathrm{mm}$ of $\mathrm{Hg}$, of an aqueous solution obtained by adding $18 \mathrm{~g}$ of glucose $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$ to $180 \mathrm{~g}$ of water at $100^{\circ} \mathrm{C}$ is
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Verified Answer
The correct answer is:
$752.4$
According to Raoult's law
$$
\frac{p^{\circ}-p_s}{p^{\circ}}=\frac{n_2}{n_1+n_2}
$$
where, $p^{\circ}=$ vapour pressure of pure water at $100^{\circ} \mathrm{C}$ $=760 \mathrm{mmHg}$.
$p_s=$ vapour pressure of solution at $100^{\circ} \mathrm{C}$
$$
\begin{aligned}
& n_2=\text { moles of solute }=\frac{W_2}{M_2}=\frac{18}{180}=0.1 \mathrm{~mol} \\
& n_1=\text { moles of solvent }=\frac{W_1}{M_1}=\frac{180}{18}=10 \mathrm{~mol}
\end{aligned}
$$
By putting these values in the formula
$$
\frac{p^{\circ}-p_s}{p^{\circ}}=\frac{0.1}{10+0.1}
$$
or
$$
\begin{aligned}
10.1\left(p^{\circ}-p_s\right) & =0.1 p^{\circ} \\
10 p^{\circ} & =10.1 p_s \\
p_s=\frac{10 \times 760}{10.1} & =752.4 \mathrm{mmHg} .
\end{aligned}
$$
$$
\frac{p^{\circ}-p_s}{p^{\circ}}=\frac{n_2}{n_1+n_2}
$$
where, $p^{\circ}=$ vapour pressure of pure water at $100^{\circ} \mathrm{C}$ $=760 \mathrm{mmHg}$.
$p_s=$ vapour pressure of solution at $100^{\circ} \mathrm{C}$
$$
\begin{aligned}
& n_2=\text { moles of solute }=\frac{W_2}{M_2}=\frac{18}{180}=0.1 \mathrm{~mol} \\
& n_1=\text { moles of solvent }=\frac{W_1}{M_1}=\frac{180}{18}=10 \mathrm{~mol}
\end{aligned}
$$
By putting these values in the formula
$$
\frac{p^{\circ}-p_s}{p^{\circ}}=\frac{0.1}{10+0.1}
$$
or
$$
\begin{aligned}
10.1\left(p^{\circ}-p_s\right) & =0.1 p^{\circ} \\
10 p^{\circ} & =10.1 p_s \\
p_s=\frac{10 \times 760}{10.1} & =752.4 \mathrm{mmHg} .
\end{aligned}
$$
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