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\(160 \mathrm{~g}\) of non-volatile solute ' \(A\) ' is dissolved in \(54 \mathrm{~mL}\) of water at \(373 \mathrm{~K}\). What is the vapour pressure of aqueous solution of \(A\). (Given, molecular weight of \(A=160 \mathrm{~g} \mathrm{~mol}^{-1}\) )
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The correct answer is:
570 Torr
RLVP \(=\frac{\left(p^{\circ}-p\right)}{p^{\circ}}\)
\(\left[\because p^{\circ}=\right.\) vapour pressure of \(\mathrm{H}_2 \mathrm{O}(g)\) at \(373 \mathrm{~K}=760\) bar
Let, \(d\) water \(=1 \mathrm{~g} / \mathrm{mL}\)
\(\Rightarrow 54 \mathrm{~mL}=54 \mathrm{~g}\) water
\(\because \chi \rightarrow\) mole fraction
\(n \rightarrow\) mole]
\(\begin{aligned}
\Rightarrow & \chi_A & =\frac{n_A}{n_A+n_{\mathrm{H}_2 \mathrm{O}}} \\
\Rightarrow & \frac{760-p}{760} & =\frac{\frac{160}{160}}{\frac{160}{160}+\frac{54}{18}}=\frac{1}{4} \\
\Rightarrow & p & =\frac{3}{4} \times 760=570 \text { tour }
\end{aligned}\)
\(\left[\because p^{\circ}=\right.\) vapour pressure of \(\mathrm{H}_2 \mathrm{O}(g)\) at \(373 \mathrm{~K}=760\) bar
Let, \(d\) water \(=1 \mathrm{~g} / \mathrm{mL}\)
\(\Rightarrow 54 \mathrm{~mL}=54 \mathrm{~g}\) water
\(\because \chi \rightarrow\) mole fraction
\(n \rightarrow\) mole]
\(\begin{aligned}
\Rightarrow & \chi_A & =\frac{n_A}{n_A+n_{\mathrm{H}_2 \mathrm{O}}} \\
\Rightarrow & \frac{760-p}{760} & =\frac{\frac{160}{160}}{\frac{160}{160}+\frac{54}{18}}=\frac{1}{4} \\
\Rightarrow & p & =\frac{3}{4} \times 760=570 \text { tour }
\end{aligned}\)
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