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NVAEasyJEE 2024Raoult's Law & Vapour Pressure

JEE Chemistry 2024 Question with Solution

The vapor pressure of pure benzene and methyl benzene at 27 °C27 \, \text{°C} is given as 80 Torr80 \, \text{Torr} and 24 Torr24 \, \text{Torr}, respectively. The mole fraction of methyl benzene in the vapor phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is _____ ×10−2\times 10^{-2} (nearest integer).

Answer

Correct answer:23

Step-by-step solution

Standard Method

Given: Pure benzene has vapor pressure 80 Torr80 \, \text{Torr} and pure methyl benzene has vapor pressure 24 Torr24 \, \text{Torr} at 27 °C27 \, \text{°C}. The liquid mixture is equimolar, so the liquid-phase mole fractions are xbenzene=0.5x_{\text{benzene}} = 0.5 and xmethylbenzene=0.5x_{\text{methylbenzene}} = 0.5.

Find: The mole fraction of methyl benzene in the vapor phase in the form _____×10−2\_\_\_\_\_ \times 10^{-2}.

For an ideal solution, apply Raoult’s law to each component:

Pbenzene=xbenzenePbenzenepureP_{\text{benzene}} = x_{\text{benzene}} P^{\text{pure}}_{\text{benzene}} Pmethylbenzene=xmethylbenzenePmethylbenzenepureP_{\text{methylbenzene}} = x_{\text{methylbenzene}} P^{\text{pure}}_{\text{methylbenzene}}

Substituting the given values:

Pbenzene=0.5×80=40 TorrP_{\text{benzene}} = 0.5 \times 80 = 40 \, \text{Torr} Pmethylbenzene=0.5×24=12 TorrP_{\text{methylbenzene}} = 0.5 \times 24 = 12 \, \text{Torr}

Using Dalton’s law, total vapor pressure is:

Ptotal=Pbenzene+Pmethylbenzene=40+12=52 TorrP_{\text{total}} = P_{\text{benzene}} + P_{\text{methylbenzene}} = 40 + 12 = 52 \, \text{Torr}

Now the mole fraction of methyl benzene in the vapor phase is:

Ymethylbenzene=PmethylbenzenePtotal=1252≈0.2308Y_{\text{methylbenzene}} = \frac{P_{\text{methylbenzene}}}{P_{\text{total}}} = \frac{12}{52} \approx 0.2308

Expressing this as ×10−2\times 10^{-2}:

0.2308=23.08×10−20.2308 = 23.08 \times 10^{-2}

Rounding to the nearest integer gives 2323. Therefore, the required numerical answer is 23.

Direct Ratio Method

Given: The mixture is equimolar, so both liquid-phase mole fractions are equal.

Find: Vapor-phase mole fraction of methyl benzene.

Because both components have the same liquid-phase mole fraction 0.50.5, that common factor cancels when forming the vapor-phase mole fraction:

Ymethylbenzene=0.5×240.5×80+0.5×24=2480+24Y_{\text{methylbenzene}} = \frac{0.5 \times 24}{0.5 \times 80 + 0.5 \times 24} = \frac{24}{80 + 24} Ymethylbenzene=24104≈0.2308=23.08×10−2Y_{\text{methylbenzene}} = \frac{24}{104} \approx 0.2308 = 23.08 \times 10^{-2}

Hence, the nearest integer is 23.

Common mistakes

  • Using the pure vapor pressure 24 Torr24 \, \text{Torr} directly as the vapor-phase mole fraction is incorrect. That value is only the pure-component vapor pressure. First find the partial pressure in solution using Raoult’s law, then divide by total pressure.

  • Forgetting that the liquid mixture is equimolar leads to wrong partial pressures. Since the mixture is equimolar, both liquid mole fractions are 0.50.5, not 11.

  • Calculating partial pressures correctly but not summing them to get total pressure gives an incorrect vapor-phase mole fraction. Use Dalton’s law and take the ratio PmethylbenzenePtotal\frac{P_{\text{methylbenzene}}}{P_{\text{total}}}.

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