JEE Main 2024 Chemistry question

ChemistrySolutionsJEE Main 2024

2.5 g of a non-volatile, non-electrolyte is dissolved in 100 g of water at 25°C. The solution showed a boiling point elevation by 2°C. Assuming the solute concentration in negligible with respect to the solvent concentration, the vapour pressure of the resulting aqueous solution is mm of Hg (nearest integer) [Given: Molal boiling point elevation constant of water () = 0.52 K·kg mol, 1 atm pressure = 760 mm of Hg, molar mass of water = 18 g mol]

Answer: 707

Solution

m = 2/0.52 = 3.85 mol kg⁻¹, so 0.385 mol solute in 5.56 mol water: ≈ n₂/n₁ = 0.0692 and $p

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Question ID VK-011455