Solutions JEE 2018: Raoult's Law and Liquid Mole Ratio

What liquid composition does the JEE Advanced 2018 question ask for?

The required liquid-phase ratio is 19 in this JEE Advanced 2018 Solutions question from Chemistry, Paper 1. It is a numerical-answer problem with no options, not a JEE Main question or a solution set for the entire paper.

Liquids A and B mix ideally at every composition. At temperature T, their equal-mole mixture has total vapour pressure 45 Torr, while pure A has vapour pressure 20 Torr. A second mixture at the same temperature has total vapour pressure 22.5 Torr. Find its liquid-phase ratio:

\frac{x_{A}}{x_{B}}

The question bank rates it medium, with an expected solving time of 90 seconds. This is the question bank’s estimate, not an official exam time limit.

How does the equimolar mixture give pure B’s vapour pressure?

Pure B’s vapour pressure is 70 Torr. The official method starts with the equimolar mixture: its total pressure is the average of the two pure-component vapour pressures.

Raoult’s law adds both partial pressures:

P_{T} = x_{A} P_{A}^{0} + x_{B} P_{B}^{0}

On the left is the total vapour pressure. On the right, the x values are liquid mole fractions; the superscript zero denotes each pure liquid’s vapour pressure at temperature T.

Equimolar means equal numbers of moles, not equal masses. Therefore: $x_{A} = x_{B} = \frac{1}{2}$

Substitute the first mixture’s pressure:

45 = \frac{1}{2} P_{A}^{0} + \frac{1}{2} P_{B}^{0}

Using Torr throughout:

45 = \frac{1}{2} (20) + \frac{1}{2} P_{B}^{0} = 10 + \frac{1}{2} P_{B}^{0}

Subtract 10, then multiply by 2: $35 = \frac{1}{2} P_{B}^{0}$ $P_{B}^{0} = 70 Torr$

The reference pressures remain 20 Torr and 70 Torr for the second mixture because the temperature has not changed. Composition changes the partial pressures, not these pure-liquid reference values.

How do you calculate both liquid mole fractions and their ratio?

Use the sum of the liquid mole fractions to eliminate B’s fraction, then solve for A’s. The requested A-to-B ratio is 19; keep A in the numerator to avoid reversing the answer.

The official substitution route starts with: $x_{A} + x_{B} = 1$ $x_{B} = 1 - x_{A}$

At the unchanged temperature, Raoult’s law gives:

P_{T} = x_{A} P_{A}^{0} + x_{B} P_{B}^{0}

Insert the second mixture’s pressure and both reference pressures:

22.5 = 20 x_{A} + 70 (1 - x_{A})

Expand the bracket, keeping the negative sign on the term containing A’s mole fraction: $22.5 = 20 x_{A} + 70 - 70 x_{A}$

Collect terms: $22.5 = 70 - 50 x_{A}$ $50 x_{A} = 47.5$

x_{A} = \frac{47.5}{50} = \frac{19}{20}

Find B’s mole fraction by subtraction, not by taking A’s reciprocal:

x_{B} = 1 - \frac{19}{20} = \frac{1}{20}

Now divide in the order asked:

\frac{x_{A}}{x_{B}} = \frac{\frac{19}{20}}{\frac{1}{20}} = 19

The final numerical answer is 19. The ratio is dimensionless, so do not attach Torr to it.

Why must the low-pressure mixture be rich in A?

The total pressure is a mole-fraction-weighted average of 20 Torr and 70 Torr, so it must lie between them. Since 22.5 Torr is close to 20 Torr, the liquid must contain much more of the less volatile component, A. This supports a large A-to-B ratio, not its inverse.

Check the partial pressures:

p_{A} = \frac{19}{20} (20) = 19 Torr

p_{B} = \frac{1}{20} (70) = 3.5 Torr

Their sum reproduces the given pressure:

p_{A} + p_{B} = 19 + 3.5 = 22.5 Torr

The calculated fractions confirm that A dominates the liquid. They also sum to one:

x_{A} = 0.95, x_{B} = 0.05

x_{A} + x_{B} = 0.95 + 0.05 = 1

Why is treating A as non-volatile wrong?

Less volatile does not mean non-volatile: pure A has a stated vapour pressure of 20 Torr. Both liquids contribute to the total pressure, so deleting A’s contribution gives the wrong ratio.

The following is an illustrative wrong calculation, not an official distractor. The supplied question is numerical-answer, with no options.

Start incorrectly with: $22.5 = 70 x_{B}$

This leads to:

x_{B} = \frac{22.5}{70} = \frac{9}{28}

x_{A} = 1 - \frac{9}{28} = \frac{19}{28}

\frac{x_{A}}{x_{B}} = \frac{19 / 28}{9 / 28} = \frac{19}{9}

The arithmetic follows from the starting equation, but that equation silently removes A’s partial pressure. The single-contributor equation would be: $P_{T} = x_{B} P_{B}^{0}$

This would require A’s vapour contribution to be negligible, which is not the stated model. In the correct solution, A supplies 19 Torr out of 22.5 Torr, so omitting it is not a harmless approximation.

Which three practice questions test the same method?

These three original practice variations test total pressure, liquid composition and vapour composition. They are not additional verified JEE PYQs. Use the same ideal A–B system at temperature T throughout:

P_{A}^{0} = 20 Torr, P_{B}^{0} = 70 Torr

What is the total pressure when A’s liquid mole fraction is three-fifths?

The total pressure is 40 Torr. Find B’s fraction by subtraction, then add both partial pressures:

x_{A} = \frac{3}{5}, x_{B} = 1 - \frac{3}{5} = \frac{2}{5}

P_{T} = \frac{3}{5} (20) + \frac{2}{5} (70) = 12 + 28 = 40 Torr

What is the liquid A-to-B ratio when total pressure is 30 Torr?

The required ratio is 4. Follow the same substitution route as the original question, changing only the total pressure: $30 = 20 x_{A} + 70 (1 - x_{A})$ $30 = 70 - 50 x_{A}$ $50 x_{A} = 40$

x_{A} = \frac{4}{5}, x_{B} = 1 - \frac{4}{5} = \frac{1}{5}

\frac{x_{A}}{x_{B}} = \frac{4 / 5}{1 / 5} = 4

What is A’s vapour mole fraction in the original 22.5 Torr mixture?

Divide A’s partial pressure by the total vapour pressure, not by either pure-liquid reference pressure. Dalton’s law gives:

y_{A} = \frac{p_{A}}{P_{T}}

Using the partial pressure already checked:

y_{A} = \frac{19}{22.5} = \frac{38}{45}

This is not A’s liquid mole fraction:

y_{A} = \frac{38}{45} \neq x_{A} = \frac{19}{20}

Raoult’s law uses liquid composition; Dalton’s law gives vapour composition. B is more volatile, so the vapour contains a larger proportion of B than the liquid does.

Use the Free JEE Chemistry Question Bank: PYQs, PDFs and Mocks for further chapter practice. Before substituting in your next question, mark whether the requested composition belongs to the liquid or vapour.

Frequently asked questions

What is the answer to the JEE Advanced 2018 ideal solutions question?

The required liquid-phase ratio x_A/x_B is 19. Applying Raoult's law to the 22.5 Torr mixture gives x_A = 19/20 and x_B = 1/20. The ratio is dimensionless, so the numerical answer has no unit.

How do you find pure B's vapour pressure from the equimolar mixture?

An equimolar mixture has liquid mole fractions x_A = x_B = 1/2. Raoult's law therefore gives 45 = (20 + P_B^0)/2, so pure B's vapour pressure is 70 Torr. Equimolar means equal moles, not equal masses.

Why can't liquid A be treated as non-volatile?

Pure A has a stated vapour pressure of 20 Torr, so it contributes to the mixture's total vapour pressure. In the 22.5 Torr mixture, A contributes 19 Torr and B contributes 3.5 Torr. Ignoring A's contribution changes the model and produces the wrong ratio.

What is A's vapour mole fraction in the 22.5 Torr mixture?

A's vapour mole fraction is y_A = p_A/P_T = 19/22.5 = 38/45, using Dalton's law. This differs from its liquid mole fraction, x_A = 19/20. Raoult's law uses liquid mole fractions to calculate partial pressures; Dalton's law converts those partial pressures into vapour mole fractions.