JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Solutions

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What weight of glucose must be dissolved in 100 g of water to lower the vapour pressure by 0.20 mm Hg ?
(Assume dilute solution is being formed)
Given : Vapour pressure of pure water is 54.2 mm Hg at room temperature. Molar mass of glucose is

180 g mol^{- 1}

(A)4.69 g
(B)3.59 g
(C)2.59 g
(D)3.69 g✓ Correct

Answer: (D)

Solution

Using Raoult's law for dilute solutions:

\frac{P ^{0} - P _{s}}{P ^{0}} = \frac{n _{solute}}{n _{solvent}}

.

Given:

P^{0} = 54.2

mm Hg,

P^{0} - P_{s} = 0.20

mm Hg, mass of water = 100 g, so

n_{solvent} = \frac{100}{18}

mol.

\frac{0.20}{54.2} = \frac{n _{solute}}{100/18}

n_{solute} = \frac{0.20 \times 100}{54.2 \times 18} = \frac{20}{975.6}

mol.

Mass of glucose =

n \times M = \frac{20}{975.6} \times 180 \approx 3.69

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option

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