The vapour pressure of $30 \%(\mathrm{w} / \mathrm{v})$ aqueous solution of glucose is \_\_\_\_ mm Hg at $25^{\circ} \mathrm{C}$. [Given : The density of $30 \%(\mathrm{w} / \mathrm{v})$, aqueous solution of glucose is $1.2 \mathrm{~g} \mathrm{~cm}^{-3}$ and vapour pressure of pure water is 24 mm Hg .] (Molar mass of glucose is $180 \mathrm{~g} \mathrm{~mol}^{-1}$.)

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Solutions

The vapour pressure of

30% (w / v)

aqueous solution of glucose is ____ mm Hg at

2 5^{\circ} C

.
[Given : The density of

30% (w / v)

, aqueous solution of glucose is

1.2 g cm^{- 3}

and vapour pressure of pure water is 24 mm Hg .]
(Molar mass of glucose is

180 g mol^{- 1}

Answer: 23

Solution

For 100 mL solution: mass of glucose = 30 g, mass of solution =

1.2 \times 100 = 120

g, mass of water =

120 - 30 = 90

g. Molality

m = \frac{30 \times 1000}{180 \times 90} = 1.85

mol/kg. Using the relation for dilute solutions:

\frac{P ^{0} - P _{s}}{P _{s}} = \frac{m \times 18}{1000}

, we get

\frac{24 - P _{s}}{P _{s}} = \frac{1.85 \times 18}{1000} = 0.0333

. Solving:

24 - P_{s} = 0.0333 P_{s}

, so

P_{s} (1.0333) = 24

, giving

P_{s} = 23.22 \approx 23

mm Hg.

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Integer answer

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