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JEE Main 20244 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2024 · 4 April · Morning Shift · Chemistry · Solutions

(2.5 ~g of a non-volatile, non-electrolyte is dissolved in (100 ~g...

(2.5  g\mathrm{~g} of a non-volatile, non-electrolyte is dissolved in (100  gof\mathrm{~g} ofwater at (25^{{}^{\circ}}C.The\mathrm{C}. Thesolution showed a boiling point elevation by (2^{{}^{\circ}}C.Assumingthe\mathrm{C}. Assuming thesolute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is (\qquad) (mm\mathrm{mm} of (Hg\mathrm{Hg} (nearest integer) [Given : Molal boiling point elevation constant of water ((Kb)=\left(\mathrm{K}_{\mathrm{b}}\right)=0.52  K.kgmol−1\mathrm{~K} . \mathrm{kg} \mathrm{mol}^{-1}, (1  atm\mathrm{~atm} pressure (=760  mm\mathrm{~mm} of (Hg\mathrm{Hg}, molar mass of water (=18  g mol−1\mathrm{~g} \mathrm{~mol}^{-1}]

Correct answer

0

Step-by-step solution

2=0.52×mm=20.52\begin{aligned} & 2=0.52 \times \mathrm{m} \\ & \mathrm{m}=\frac{2}{0.52} \end{aligned}According to question, solution is much diluted  so ΔPP∘=nsolute nsolvent ΔPP∘=m1000×Msolvent ΔP=P∘×m1000×Msolvent =760×20.521000×18=52.615P5=760−52.615=707.385mm of Hg\begin{aligned} & \text{ so } \frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}=\frac{\mathrm{n}_{\text{solute }}}{\mathrm{n}_{\text{solvent }}} \\ & \frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}=\frac{\mathrm{m}}{1000} \times \mathrm{M}_{\text{solvent }} \\ & \Delta \mathrm{P}=\mathrm{P}^{\circ} \times \frac{\mathrm{m}}{1000} \times \mathrm{M}_{\text{solvent }} \\ & \quad=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}
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