JEE MAIN - Chemistry (2022 - 27th July Evening Shift - No. 17)
When a certain amount of solid A is dissolved in $$100 \mathrm{~g}$$ of water at $$25^{\circ} \mathrm{C}$$ to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is $$23.76 \,\mathrm{mmHg}$$. The number of moles of solute A added is _____________. (Nearest Integer)
Answer
6
Explanation
$\because$ Diliute solution given:
$$ \frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^0} \sim \frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $$
$$ \frac{\mathrm{P}^0-\mathrm{P}^0 / 2}{\mathrm{P}^0}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $$
${ }^{\mathrm{n}}$ solute $\sim \frac{{ }^{\mathrm{n}} \text { solvent }}{2}=\frac{100}{18 \times 2}=2.78 \mathrm{~mol}$
More accurate approach:
$$ \frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_{\mathrm{S}}}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $$
$$ \frac{\mathrm{P}^0-\mathrm{P}^0 / 2}{\mathrm{P}^0 / 2}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^n \text { solvent }} $$
${ }^n$ solute $={ }^n$ solvent $=\frac{100}{18}=5.55 \mathrm{~mol}$
$$ \frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^0} \sim \frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $$
$$ \frac{\mathrm{P}^0-\mathrm{P}^0 / 2}{\mathrm{P}^0}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $$
${ }^{\mathrm{n}}$ solute $\sim \frac{{ }^{\mathrm{n}} \text { solvent }}{2}=\frac{100}{18 \times 2}=2.78 \mathrm{~mol}$
More accurate approach:
$$ \frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_{\mathrm{S}}}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $$
$$ \frac{\mathrm{P}^0-\mathrm{P}^0 / 2}{\mathrm{P}^0 / 2}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^n \text { solvent }} $$
${ }^n$ solute $={ }^n$ solvent $=\frac{100}{18}=5.55 \mathrm{~mol}$
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