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JEE MAINS 2024
09.04.24 S1
Question
2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at $25^{\circ} \mathrm{C}$. The solution showed boiling point elevation by $2^{\circ} \mathrm{C}$. Assuming the solute concentration in negligible with respect to the solvent concentration, the vapour pressure of the resulting aqueous solution is $\_\_\_\_$ mm of (nearest integer)| [Givedi Molal boiling point elevation constant of water $\left(\mathrm{K}_{\mathrm{b}}\right)=0.52 \mathrm{~K} . \mathrm{kgmol}^{-1}$, 1 atm pressure $=760 \mathrm{~mm}$ of Hg , molar mass of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
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Solution
Sol. $$ \begin{aligned} & 2=0.52 \times \mathrm{m} \\ & \mathrm{~m}=\frac{2}{0.52} \end{aligned} $$ According to question, solution is much diluted $\begin{aligned} & \text { so } \frac{\Delta \mathrm{P}}{\mathrm{P}^0}=\frac{\mathrm{n}_{\text {solute }}}{\mathrm{n}_{\text {solvent }}} \\ & \frac{\Delta \mathrm{P}}{\mathrm{P}^0}=\frac{\mathrm{m}}{1000} \times \mathrm{M}_{\text {solvent }} \\ & \Delta \mathrm{P}=\mathrm{P}^0 \times \frac{\mathrm{m}}{1000} \times \mathrm{M}_{\text {solvent }} \\ & =760 \times \frac{\frac{2}{0.52}}{1000} \times 18=52.615 \\ & \mathrm{P}_5=760-52.615=707.385 \mathrm{~mm} \text { of } \mathrm{Hg}\end{aligned}$
Question Tags
JEE Main
Chemistry
Medium
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