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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\degree \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 vapor pressure =80 Torr= 80 \, \text{Torr}, pure methyl benzene vapor pressure =24 Torr= 24 \, \text{Torr}, and the liquid mixture is equimolar, so 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, expressed as _____ ×10−2\times 10^{-2}.

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

Pbenzene=Xbenzene⋅PbenzenepureP_{\text{benzene}} = X_{\text{benzene}} \cdot P^{\text{pure}}_{\text{benzene}} Pmethylbenzene=Xmethylbenzene⋅PmethylbenzenepureP_{\text{methylbenzene}} = X_{\text{methylbenzene}} \cdot 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}

Now use Dalton’s law to find the total vapor pressure:

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

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 in the form _____ ×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 value is 2323.

Direct Ratio Shortcut

Given: The liquid mixture is equimolar.

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

Because both liquid mole fractions are equal, the common factor 0.50.5 appears in both partial pressures and cancels in the vapor-phase ratio:

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}

So,

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 2323.

Common mistakes

  • Using the pure vapor pressure of methyl benzene directly as its vapor-phase mole fraction is incorrect. 24 Torr24 \, \text{Torr} is only the pure-component vapor pressure. First find its partial pressure in the equimolar liquid mixture using Raoult’s law, then divide by the total pressure.

  • Forgetting that an equimolar liquid mixture means Xbenzene=Xmethylbenzene=0.5X_{\text{benzene}} = X_{\text{methylbenzene}} = 0.5 leads to wrong partial pressures. Do not use 11 for each component; use the liquid-phase mole fractions correctly.

  • Calculating the ratio as 2480+24\frac{24}{80+24} without understanding why can cause errors in other questions. This simplification works here only because both liquid mole fractions are equal and cancel out. In general, partial pressures must be computed first.

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