JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Solutions

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A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of A=100 g mol⁻¹; B= 200 g mol⁻¹; C=10,000 g mol⁻¹]

(A)B > C > A
(B)C > B > A
(C)A > B > C✓ Correct
(D)A > C > B

Answer: (C)

Solution

Relative lowering of vapour pressure equals the mole fraction of solute, and for dilute solutions it runs in proportion to the moles of solute since the solvent amount is fixed. With

10 g

of each solute, the moles are

n_{A} = \frac{10}{100} = 0.1

n_{B} = \frac{10}{200} = 0.05

and

n_{C} = \frac{10}{10000} = 0.001

. As the effect tracks the number of moles, the lowering ranks as

A > B > C

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

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