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← JEE Advanced 2022 · Paper 2
ChemistryInteger+3 / -1Question 38 of 54

JEE Advanced 2022 · Paper 2, Chemistry question 2

An aqueous solution is prepared by dissolving 0.1 mol of an ionic salt in 1.8 kg of water at 35 ºC. The salt remains 90% dissociated in the solution. The vapour pressure of the solution is 59.724 mm of Hg. Vapor pressure of water at 35 ºC is 60.000 mm of Hg. The number of ions present per formula unit of the ionic salt is ________.

Answer and worked solution

Official answer: 5

Relative lowering of vapour pressure.
p0−pp0=0.27660=0.0046.\frac{p^0-p}{p^0}=\frac{0.276}{60}=0.0046.

Van 't Hoff factor. There are 180018=100\frac{1800}{18}=100 mol of water, so
0.1 i100+0.1 i=0.0046  ⟹  i≈4.6.\frac{0.1\,i}{100+0.1\,i}=0.0046\implies i\approx4.6.

Ions per formula unit. With 90% dissociation, i=1+0.9(n−1)=4.6i=1+0.9(n-1)=4.6, so n−1=4n-1=4 and n=5n=5.

Topics:van't Hoff factorRaoult's law and vapour pressure
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