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Solutions question

2024 · 4 Apr · Shift 1 · Q21
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Solutions question

2024 · 4 Apr · Shift 1 · Q21

JEE MainChemistrySolutionsNumerical+4 / −1
2.5 g2.5 \mathrm{~g}2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g100 \mathrm{~g}100 g of water at 25∘C25^{\circ} \mathrm{C}25∘C. The solution showed a boiling point elevation by 2∘C2^{\circ} \mathrm{C}2∘C. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is ‾mm\underline{\hspace{2cm}}\mathrm{mm}​mm of Hg\mathrm{Hg}Hg(nearest integer) [Given : Molal boiling point elevation constant of water (Kb)=0.52 K.kgmol−1\left(\mathrm{K}_{\mathrm{b}}\right)=0.52 \mathrm{~K} . \mathrm{kg} \mathrm{mol}^{-1}(Kb​)=0.52 K.kgmol−1, 1 atm1 \mathrm{~atm}1 atm pressure =760 mm=760 \mathrm{~mm}=760 mm of Hg\mathrm{Hg}Hg, molar mass of water =18 g mol−1]=18 \mathrm{~g} \mathrm{~mol}^{-1}]=18 g mol−1]
Numerical answer
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