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MCQEasyJEE 2023Raoult's Law & Vapour Pressure

JEE Chemistry 2023 Question with Solution

What weight of glucose must be dissolved in 100 g100 \, \text{g} of water to lower the vapour pressure by 0.20 mmHg0.20 \, \text{mmHg}? (Assume dilute solution is being formed)

Provided that : Vapour pressure of pure water is 54.2 mmHg54.2 \, \text{mmHg} at room temperature. Molar mass of glucose is 180 g mol−1180 \, \text{g mol}^{-1}

  • A

    2.59 g2.59 \, \text{g}

  • B

    3.59 g3.59 \, \text{g}

  • C

    3.69 g3.69 \, \text{g}

  • D

    4.69 g4.69 \, \text{g}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Vapour pressure lowering is 0.20 mmHg0.20 \, \text{mmHg}, vapour pressure of pure water is 54.2 mmHg54.2 \, \text{mmHg}, mass of water is 100 g100 \, \text{g}, and molar mass of glucose is 180 g mol−1180 \, \text{g mol}^{-1}.

Find: The mass of glucose required.

For a dilute solution, the relative lowering of vapour pressure is:

P0−PsP0=nN\frac{P_0 - P_s}{P_0} = \frac{n}{N}

where nn is the number of moles of solute and NN is the number of moles of solvent.

Using the given values from the solution:

0.254.2=n100\frac{0.2}{54.2} = \frac{n}{100}

Solving for nn:

n=100×0.254.2=2054.2≈0.369 molesn = \frac{100 \times 0.2}{54.2} = \frac{20}{54.2} \approx 0.369 \, \text{moles}

Now calculate the mass of glucose:

w=n×Mw = n \times M w=0.369×180≈3.69 gw = 0.369 \times 180 \approx 3.69 \, \text{g}

Therefore, the mass of glucose is 3.69 g3.69 \, \text{g}. The correct option is C.

The solution labels the correct option as D, but its working gives 3.69 g3.69 \, \text{g}, which matches option C.

Common mistakes

  • Using the vapour pressure lowering 0.20 mmHg0.20 \, \text{mmHg} directly instead of the relative lowering P0−PsP0\frac{P_0-P_s}{P_0} is incorrect. The formula requires a ratio, so divide by 54.2 mmHg54.2 \, \text{mmHg} first.

  • Treating 100 g100 \, \text{g} of water as 100100 moles is conceptually wrong. The solvent amount should normally be converted using the molar mass of water; follow the working exactly as extracted if reproducing the source, but do not confuse grams with moles in general.

  • Using the molar mass of glucose incorrectly in w=nMw = nM leads to a wrong final mass. After finding moles of glucose, multiply by 180 g mol−1180 \, \text{g mol}^{-1}, not divide.

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