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The vapour pressure of water at $23^{\circ} \mathrm{C}$ is $19.8 \mathrm{~mm}, 0.1$ mole of glucose is dissolved in $178.2 \mathrm{~g}$ of water. What is the vapour pressure (in $\mathrm{mm}$ ) of the resultant solution?
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Verified Answer
The correct answer is:
19.602
Given $P_s=19.8 \mathrm{~mm}$
$\begin{aligned}
& n_A=0.1 \\
& n_B=\frac{178.2}{18}=9.9
\end{aligned}$
According to Raoult's law
$\begin{gathered}
\frac{P_s-P}{P_s}=\frac{n_A}{n_A+n_B} \\
\frac{19.8-P}{19.8}=\frac{0.1}{9.9+0.1}
\end{gathered}$
or
$\begin{aligned}
198-10 P & =19.8 \times 0.1 \\
10 P & =198-1.98 \\
10 P & =196.02 \\
P & =19.602 \mathrm{~mm}
\end{aligned}$
$\begin{aligned}
& n_A=0.1 \\
& n_B=\frac{178.2}{18}=9.9
\end{aligned}$
According to Raoult's law
$\begin{gathered}
\frac{P_s-P}{P_s}=\frac{n_A}{n_A+n_B} \\
\frac{19.8-P}{19.8}=\frac{0.1}{9.9+0.1}
\end{gathered}$
or
$\begin{aligned}
198-10 P & =19.8 \times 0.1 \\
10 P & =198-1.98 \\
10 P & =196.02 \\
P & =19.602 \mathrm{~mm}
\end{aligned}$
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