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The vapour pressure of pure water is $23 \mathrm{mmHg}$. The vapour pressure of an aqueous solution, which contains 10 mass per cent of solute ' $A$ ' having molecular weight 50 is
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$0.028 \mathrm{~atm}$
Given,
Vapour pressure of pure water $=23 \mathrm{~mm} \mathrm{Hg} 10 \%$ of solute $A$
Molecular weight of $A=50$
Mass of $A=10 \mathrm{~g}$
Mass of water $=100-10=90 \mathrm{~g}$
Number of moles of $A=\frac{10}{50}=0.2$
Number of moles of water $=\frac{90}{18}=5$
$\therefore$ Total number of moles $=5+0.2=5.2$
Mole fraction of solvent $=\frac{5}{5.2}$
$p_s=$ mole fraction of solvent $\times p^{\circ}$
$\frac{5}{2.5} \times 23=22.11 \mathrm{~mm} \mathrm{Hg} \approx 22 \mathrm{~mm} \mathrm{Hg}$
$=\frac{22}{760}=0.028 \mathrm{~atm}$.
Vapour pressure of pure water $=23 \mathrm{~mm} \mathrm{Hg} 10 \%$ of solute $A$
Molecular weight of $A=50$
Mass of $A=10 \mathrm{~g}$
Mass of water $=100-10=90 \mathrm{~g}$
Number of moles of $A=\frac{10}{50}=0.2$
Number of moles of water $=\frac{90}{18}=5$
$\therefore$ Total number of moles $=5+0.2=5.2$
Mole fraction of solvent $=\frac{5}{5.2}$
$p_s=$ mole fraction of solvent $\times p^{\circ}$
$\frac{5}{2.5} \times 23=22.11 \mathrm{~mm} \mathrm{Hg} \approx 22 \mathrm{~mm} \mathrm{Hg}$
$=\frac{22}{760}=0.028 \mathrm{~atm}$.
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