Solutions JEE 2024: Boiling-Point Elevation Ratio Explained

What is the answer to the two-vessel Solutions JEE 2024 question?

The answer to this Solutions JEE 2024 question is 150, not 50: it asks for one boiling-point elevation as a percentage of the other, not the percentage increase. This is a JEE Advanced 2024, Paper 2, Chemistry numerical-answer question from Solutions, not an MCQ.

Vessel 1 contains non-volatile solute X; Vessel 2 contains non-volatile solute Y. Each has the following solute and water masses, at the same temperature and pressure:

Solute mass = w_{2} g, water mass = w_{1} g

At their respective concentrations:

M_{X} = 0.8 M_{Y}, i_{X} = 1.2 i_{Y}

Find Vessel 1’s boiling-point elevation as a percentage of Vessel 2’s. On the question bank’s scale, this is medium; its 90 seconds solving-time estimate is not an official exam time limit.

How do you write the boiling-point elevation for each vessel?

For a dilute solution, use the boiling-point elevation relation below. Write a separate equation for each vessel before taking the ratio: this keeps the molar masses in the correct positions. $Δ T_{b} = i K_{b} m$

The symbols represent boiling-point elevation, van’t Hoff factor, the solvent’s ebullioscopic constant and molality, respectively. Molality means moles of solute divided by kilograms of solvent, not kilograms of solution:

m = \frac{moles of solute}{mass of solvent in kg}

Convert the water mass before substituting. Do not add the solute mass to this denominator.

w_{1} g o f w a t e r = \frac{w_{1}}{1000} k g o f w a t e r

With molar masses expressed in grams per mole, the solute amounts are:

n_{X} = \frac{w_{2}}{M_{X}}, n_{Y} = \frac{w_{2}}{M_{Y}}

The official method gives these two elevations:

(Δ T_{b})_{1} = i_{X} K_{b} [\frac{w_{2} / M_{X}}{w_{1} / 1000}]

(Δ T_{b})_{2} = i_{Y} K_{b} [\frac{w_{2} / M_{Y}}{w_{1} / 1000}]

The ebullioscopic constant is common because both solvents are water. Equal solute masses do not mean equal molalities: different molar masses give different numbers of moles.

What cancels when you divide the two elevation equations?

The solvent constant, solute mass and solvent mass cancel, leaving the van’t Hoff factor ratio multiplied by the inverse molar-mass ratio. The molar mass of Y belongs in the numerator. First divide the complete expressions:

\frac{(Δ T_{b})_{1}}{(Δ T_{b})_{2}} = \frac{i_{X} K_{b} [\frac{w_{2} / M_{X}}{w_{1} / 1000}]}{i_{Y} K_{b} [\frac{w_{2} / M_{Y}}{w_{1} / 1000}]}

Separate the factors to expose every cancellation:

= \frac{i_{X}}{i_{Y}} \underset{1}{\underset{⏟}{\frac{K_{b}}{K_{b}}}} \underset{1}{\underset{⏟}{\frac{w_{2}}{w_{2}}}} \frac{M_{Y}}{M_{X}} \underset{1}{\underset{⏟}{\frac{w_{1} / 1000}{w_{1} / 1000}}} = \frac{i_{X}}{i_{Y}} \frac{M_{Y}}{M_{X}}

Substitute the given van’t Hoff factor ratio, followed by the molar-mass relation:

\frac{i_{X}}{i_{Y}} = 1.2

M_{X} = 0.8 M_{Y} \Rightarrow \frac{M_{Y}}{M_{X}} = \frac{1}{0.8} = 1.25

Therefore:

\frac{(Δ T_{b})_{1}}{(Δ T_{b})_{2}} = 1.2 \times 1.25 = 1.5

No absolute value of either molar mass or either van’t Hoff factor is needed. The solvent constant also cancels because both solutions use water.

Why is the numerical answer 150 rather than 50?

The requested quantity is “percent of”, so multiply the elevation ratio by 100 without subtracting anything. Vessel 1’s elevation is one and a half times Vessel 2’s:

\frac{(Δ T_{b})_{1}}{(Δ T_{b})_{2}} \times 100 = 1.5 \times 100 = 150 %

Numerical entry: 150.

“150% of Vessel 2’s elevation” means “50% greater than Vessel 2’s elevation.” The percentage increase uses a different calculation: $(1.5 - 1) \times 100 = 50 %$

The comparison concerns elevations above water’s boiling point. It does not compare the absolute boiling temperatures of the two solutions.

Why does multiplying by 0.8 give the wrong answer?

Multiplying by 0.8 reverses the molar-mass ratio: the molality ratio needs the molar mass of Y divided by that of X. The supplied question has no answer options, so 96 is an illustrative wrong numerical entry, not an official distractor.

The incorrect calculation is: $1.2 \times 0.8 \times 100 = 96$

The precise error is using:

\frac{M_{X}}{M_{Y}} instead of \frac{M_{Y}}{M_{X}}

At fixed solute mass, moles depend inversely on molar mass:

n = \frac{w_{2}}{M}

X has the lower molar mass, so the same mass supplies more moles. With equal water masses, its molality is higher.

Check the direction before accepting an answer. X has both more moles at equal mass and a larger van’t Hoff factor, so its boiling-point elevation must exceed Y’s. An answer of 96% says it is smaller and fails this check.

How do you apply the ratio method to related Solutions questions?

Keep the solute-to-solvent calculation unchanged, then use the factor and molar-mass ratios for the stated conditions. The following are original practice variations, not additional verified JEE PYQs.

What if the two van’t Hoff factors are equal?

Vessel 1’s boiling-point elevation is 125% of Vessel 2’s. This variation isolates the molar-mass effect by removing the difference in van’t Hoff factors.

Question 1: Two dilute aqueous solutions contain equal masses of non-volatile solutes X in Vessel 1 and Y in Vessel 2, in equal masses of water. Given:

M_{X} = 0.8 M_{Y}, i_{X} = i_{Y}

Express Vessel 1’s boiling-point elevation as a percentage of Vessel 2’s.

Worked answer: Divide the boiling-point elevation equations and cancel the common solvent and mass factors:

\frac{(Δ T_{b})_{1}}{(Δ T_{b})_{2}} = \frac{i_{X}}{i_{Y}} \frac{M_{Y}}{M_{X}} = 1 \times 1.25 = 1.25

Requested percentage = 1.25 \times 100 = 125 %

What if the question asks for freezing-point depression?

The magnitude in Vessel 1 is 150% of that in Vessel 2, provided the stated van’t Hoff ratio applies under the freezing-point conditions. This tests transfer to another colligative property.

Question 2: Consider dilute aqueous solutions of X in Vessel 1 and Y in Vessel 2, with the original equal solute masses and equal water masses. Under the freezing-point conditions, use:

M_{X} = 0.8 M_{Y}, i_{X} = 1.2 i_{Y}

Find Vessel 1’s freezing-point depression magnitude as a percentage of Vessel 2’s.

Worked answer: Start with the depression-magnitude relation: $Δ T_{f} = i K_{f} m$

Divide the complete expressions; the common solvent constant, solute mass and water-mass factors cancel:

\frac{(Δ T_{f})_{1}}{(Δ T_{f})_{2}} = \frac{i_{X} K_{f} [(w_{2} / M_{X}) / (w_{1} / 1000)]}{i_{Y} K_{f} [(w_{2} / M_{Y}) / (w_{1} / 1000)]} = \frac{i_{X}}{i_{Y}} \frac{M_{Y}}{M_{X}}

Requested percentage = 1.2 \times 1.25 \times 100 = 150 %

Do not carry a van’t Hoff ratio into new conditions without checking it. Before substituting, confirm that the factors apply at the temperature and concentrations stated.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

Why is the answer to the Solutions JEE 2024 question 150 and not 50?

Vessel 1’s boiling-point elevation is 1.5 times Vessel 2’s, and the question asks for the first elevation as a percentage of the second. Multiply 1.5 by 100 to get 150, the required numerical entry. The value 50 describes the percentage increase, which is not what the question asks.

How do you calculate the boiling-point elevation ratio for the two vessels?

Start with boiling-point elevation = i × K_b × molality for each vessel. Since both vessels contain equal solute masses and equal water masses, the ratio reduces to (i_X/i_Y) × (M_Y/M_X). Substituting the given values gives 1.2 × (1/0.8) = 1.5.

Why do we divide by 0.8 instead of multiplying by it?

At fixed solute mass, the number of moles is inversely proportional to molar mass. Since M_X = 0.8M_Y, equal masses of X and Y give a mole ratio of M_Y/M_X = 1/0.8. Equal water masses make this the molality ratio too.

Does the same ratio method work for freezing-point depression?

Yes, for dilute solutions in the same solvent, freezing-point depression magnitude follows i × K_f × molality. With equal solute masses and equal solvent masses, the ratio again reduces to (i_X/i_Y) × (M_Y/M_X). The result is 150% only if the stated van’t Hoff factor ratio applies under the freezing-point conditions.

Solutions JEE 2024: Boiling-Point Elevation Ratio