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If $p^{\circ}$ and $p$ are the vapour pressure of the pure solvent and solution and $n_{1}$ and $n_{2}$ are the moles of solute and solvent respectively in the solution then the correct relation between $p$ and $p^{\circ}$ is
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Verified Answer
The correct answer is:
$p=p^{\circ}\left[\frac{n_{2}}{n_{1}+n_{2}}\right]$
Given,
$$
\begin{array}{l}
p^{\circ}= \text {vapour pressure of pure solvent}\\
p=\text { vapour pressure of solution } \\
n_{1}=\text { moles of solute } \\
n_{2}=\text { moles of solvent }
\end{array}
$$
Thus, according to Raoult's law.
$$
\frac{p^{\circ}-p}{p^{\circ}}=x_{1}=\frac{n_{1}}{n_{1}+n_{2}}
$$
$\left(x_{1}=\right.$ mole fraction of solute)
or, $\quad 1-\frac{p}{p^{\circ}}=\frac{n_{1}}{n_{1}+n_{2}}$
$$
\frac{p}{p^{\circ}}=1-\frac{n_{1}}{n_{1}+n_{2}}
$$
$\Rightarrow$
$$
\begin{array}{c}
\frac{n_{1}+n_{2}-n_{1}}{n_{1}+n_{2}} \\
\frac{p}{p_{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}
\end{array}
$$
$\Rightarrow p=p^{\circ} \left( \frac{n_{2}}{n_{1}+n_{2}} \right)$
$$
\begin{array}{l}
p^{\circ}= \text {vapour pressure of pure solvent}\\
p=\text { vapour pressure of solution } \\
n_{1}=\text { moles of solute } \\
n_{2}=\text { moles of solvent }
\end{array}
$$
Thus, according to Raoult's law.
$$
\frac{p^{\circ}-p}{p^{\circ}}=x_{1}=\frac{n_{1}}{n_{1}+n_{2}}
$$
$\left(x_{1}=\right.$ mole fraction of solute)
or, $\quad 1-\frac{p}{p^{\circ}}=\frac{n_{1}}{n_{1}+n_{2}}$
$$
\frac{p}{p^{\circ}}=1-\frac{n_{1}}{n_{1}+n_{2}}
$$
$\Rightarrow$
$$
\begin{array}{c}
\frac{n_{1}+n_{2}-n_{1}}{n_{1}+n_{2}} \\
\frac{p}{p_{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}
\end{array}
$$
$\Rightarrow p=p^{\circ} \left( \frac{n_{2}}{n_{1}+n_{2}} \right)$
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