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)

$\because$ Diliute solution given:<br/><br/> $$ \frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^0} \sim \frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $$<br/><br/> $$ \frac{\mathrm{P}^0-\mathrm{P}^0 / 2}{\mathrm{P}^0}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $$<br/><br/> ${ }^{\mathrm{n}}$ solute $\sim \frac{{ }^{\mathrm{n}} \text { solvent }}{2}=\frac{100}{18 \times 2}=2.78 \mathrm{~mol}$<br/><br/> More accurate approach:<br/><br/> $$ \frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_{\mathrm{S}}}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $$<br/><br/> $$ \frac{\mathrm{P}^0-\mathrm{P}^0 / 2}{\mathrm{P}^0 / 2}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^n \text { solvent }} $$<br/><br/> ${ }^n$ solute $={ }^n$ solvent $=\frac{100}{18}=5.55 \mathrm{~mol}$