JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Solutions

When a certain amount of solid A is dissolved in 100 g of water at

2 5^{\circ} C

to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is 23.76 mmHg. The number of moles of solute A added is

____

. (Nearest Integer)

Answer: 3

Solution

For a dilute solution, using Raoult's law:

\frac{P ^{0} - P _{s}}{P ^{0}} \approx \frac{n _{so l u t e}}{n _{so l v e n t}}

Given:

P_{s} = \frac{P ^{0}}{2}

\frac{P ^{0} - P ^{0} /2}{P ^{0}} = \frac{n _{so l u t e}}{n _{so l v e n t}}

\frac{1}{2} = \frac{n _{so l u t e}}{n _{so l v e n t}}

Moles of water (solvent) =

\frac{100}{18} = 5.56

mol

n_{so l u t e} = \frac{5.56}{2} = 2.78 \approx 3

mol

Therefore, the answer is 3 moles.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer

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