Question

# For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2∘C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb=0.76K kg mol−1).

A

724

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B

740

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C

736

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D

718

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Solution

## The correct option is A 724 The elevation in boiling point is ΔTb=Kb.m m=molality=n2w1×1000 [n2 = Number of moles of solute, w1 = Weight of solvent in gram] ⇒2=0.76×n2100×1000 ⇒n2=519 Also, from relative lowering of vapour pressure: −Δpp0=x2=n2n1+n2≈n2n1 [∵ n1>>n2] ⇒ −ΔP=760×519×18100 = 36 mm of Hg ⇒ p = 760 – 36 = 724 mm of Hg

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