Solutions JEE 2019: Benzene Freezing-Point Depression

What is the answer to the Solutions JEE 2019 benzene question?

The supplied key records 1.02 K for the Solutions JEE 2019 benzene question, but its displayed calculation gives approximately 1.02564 K. Those are not consistent under ordinary two-decimal rounding.

This is JEE Advanced 2019, Paper 1, Chemistry, Question 2, a numerical-answer question on colligative properties. Its question-bank tags are medium difficulty and 120 seconds, not official exam classifications.

Dissolving 0.5 g of a non-volatile, non-ionic solute in 39 g of benzene lowers the vapour pressure from 650 to 640 mm Hg. Find the freezing-point depression in kelvin, using:

M_{benzene} = 78 g m o l^{- 1}, K_{f} = 5.12 K k g m o l^{- 1}

Why does the supplied solution divide by 640 instead of 650?

640 is correct because the right-hand side is the solute-to-solvent mole ratio, not the solute mole fraction. Conventional relative lowering uses pure-solvent pressure in the denominator and equals the solute mole fraction. Dividing that relation by Raoult’s law gives the equivalent expression used in the supplied solution.

Define the quantities:

p^{0} & = pure benzene vapour pressure p_{s} & = solution vapour pressure n_{1} & = moles of benzene n_{2} & = moles of solute

Use the problem’s non-ionic, non-associating treatment, so there is no change in particle count: $i = 1$

For the ideal-solution treatment, Raoult’s law and conventional relative lowering give:

\frac{p_{s}}{p^{0}} = \frac{n_{1}}{n_{1} + n_{2}}

\frac{p^{0} - p_{s}}{p^{0}} = \frac{n_{2}}{n_{1} + n_{2}}

Divide the second relation by the first:

\frac{(p^{0} - p_{s}) / p^{0}}{p_{s} / p^{0}} = \frac{n_{2} / (n_{1} + n_{2})}{n_{1} / (n_{1} + n_{2})}

\frac{p^{0} - p_{s}}{p_{s}} = \frac{n_{2}}{n_{1}}

The pressure drop is:

p^{0} - p_{s} = 650 - 640 = 10 m m H g

The pressure units cancel in either ratio. 650 pairs with mole fraction; 640 pairs with mole ratio.

How do you calculate the solute molar mass first?

The solute molar mass is 64 grams per mole. Express both mole amounts through their masses, then substitute into the pressure-to-mole-ratio relation. Keep the solute’s 0.5 g separate from the benzene’s 0.5 mol: their numerical values match, but their units do not.

Let the solute molar mass be: $M g m o l^{- 1}$

The benzene and solute mole amounts are:

n_{1} = \frac{39}{78} = 0.5 m o l

n_{2} = \frac{0.5}{M} m o l

Substitute, using masses in grams and molar masses in grams per mole:

\frac{650 - 640}{640} = \frac{0.5 / M}{39 / 78}

\frac{10}{640} = \frac{0.5 \times 78}{39 M} = \frac{39}{39 M} = \frac{1}{M}

\frac{1}{64} = \frac{1}{M}

M = 64 g m o l^{- 1}

Now calculate the solute mole amount for the freezing-point calculation:

n_{2} = \frac{0.5}{64} = 0.0078125 m o l

How do you calculate molality and report the freezing-point depression?

The molality is approximately 0.20032 moles per kilogram, giving a freezing-point depression of approximately 1.02564 K. Molality means solute moles per kilogram of solvent, not solution. Use the 39 g of benzene, not the combined 39.5 g mass.

Convert the solvent mass:

39 g = \frac{39}{1000} k g = 0.039 k g

Then calculate molality:

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

m = \frac{0.5 / 64}{39 / 1000} = \frac{0.5 \times 1000}{64 \times 39} = \frac{500}{2496} \approx 0.20032 m o l k g^{- 1}

Apply freezing-point depression without rounding the intermediate fraction:

Δ T_{f} = i K_{f} m = 1 \times 5.12 \times \frac{500}{2496} = \frac{2560}{2496} \approx 1.02564 K

The units check is:

K k g m o l^{- 1} \times m o l k g^{- 1} = K

1.02 K remains the supplied keyed answer. However, the supplied solution’s claim of ordinary two-decimal rounding is inconsistent with its arithmetic: 1.02564 K conventionally rounds to 1.03 K, not 1.02 K.

This is a precision discrepancy, not an official acceptance rule. It does not establish that a different numerical entry would receive marks.

What goes wrong if you use 650 with the same mole ratio?

Pairing 650 with the solute-to-solvent mole ratio gives an incorrect molar mass of 65 grams per mole, followed by approximately 1.01 K. This is a wrong numerical result, not an MCQ distractor: no options are supplied because this is a numerical-answer question.

The mismatched substitution is:

\frac{10}{650} = \frac{0.5 / M}{39 / 78} = \frac{1}{M}

M = 65 g m o l^{- 1}

Carrying that value through gives:

Δ T_{f} = 5.12 \times \frac{0.5 / 65}{0.039} \approx 1.00986 K \approx 1.01 K

Using 650 is valid when the right-hand side is:

\frac{n_{2}}{n_{1} + n_{2}}

This mole fraction is not exactly equal to the solute-to-solvent mole ratio. Replacing total moles with solvent moles is a dilute-solution approximation, and that approximation changes the reported hundredth here.

Use the exact pressure-to-mole-ratio relation in the supplied method. It avoids an unnecessary approximation before the final calculation.

Which two practice questions check this method?

These two original follow-up practice questions based on this PYQ check mole fraction versus mole ratio, then molality and freezing-point depression when the solute mass doubles. They are not additional verified past-paper questions. Attempt them before checking the calculations.

Question 1: For the same solution containing 0.5 g solute in 39 g benzene, with vapour pressure falling from 650 to 640 mm Hg, calculate the solute mole fraction and solute-to-benzene mole ratio.

x_{2} = \frac{10}{650} = \frac{1}{65} \approx 0.015385

\frac{n_{2}}{n_{1}} = \frac{10}{640} = \frac{1}{64} = 0.015625

The fractions differ because their denominators count total moles and solvent moles, respectively. Both ratios are dimensionless.

Question 2: Double the mass of the same solute to 1.0 g while keeping benzene at 39 g. Find molality and freezing-point depression under the same dilute-solution assumptions, using the original freezing-point constant.

m = \frac{1 / 64}{0.039} \approx 0.40064 m o l k g^{- 1}

Δ T_{f} = 5.12 \times \frac{1 / 64}{0.039} \approx 5.12 \times 0.40064 \approx 2.05128 K

Δ T_{f} \approx 2.05 K

Use the unrounded calculation, not twice the recorded 1.02 K answer. Keep intermediate fractions intact and round only the final result.

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Frequently asked questions

What is the answer to the Solutions JEE 2019 benzene question?

The supplied key records 1.02 K, but the displayed calculation gives approximately 1.02564 K. Ordinary rounding to two decimal places gives 1.03 K, not 1.02 K. This discrepancy does not establish which numerical entries would officially receive marks.

Why do we divide by 640 instead of 650 in the benzene question?

The pressure drop divided by the solution vapour pressure equals the solute-to-solvent mole ratio: 10/640 = n₂/n₁. Dividing by the pure-solvent pressure instead gives the solute mole fraction: 10/650 = n₂/(n₁ + n₂). Both relations are valid under Raoult's law, but their mole expressions cannot be interchanged.

How do I calculate the solute molar mass in the JEE 2019 benzene question?

Benzene contributes 39/78 = 0.5 mol, while the solute contributes 0.5/M mol when M is its molar mass in g/mol. Substituting into 10/640 = (0.5/M)/0.5 gives M = 64 g/mol.

Should I use 39 g or 39.5 g to calculate molality?

Use 39 g, or 0.039 kg, because molality is solute moles per kilogram of solvent, not solution. The molality is (0.5/64)/0.039 ≈ 0.20032 mol/kg. With i = 1 and Kf = 5.12 K kg/mol, the freezing-point depression is approximately 1.02564 K.

Solutions JEE 2019: Benzene Freezing-Point Depression

What is the answer to the Solutions JEE 2019 benzene question?

Why does the supplied solution divide by 640 instead of 650?

How do you calculate the solute molar mass first?

How do you calculate molality and report the freezing-point depression?

What goes wrong if you use 650 with the same mole ratio?

Which two practice questions check this method?

Frequently asked questions

What is the answer to the Solutions JEE 2019 benzene question?

Why do we divide by 640 instead of 650 in the benzene question?

How do I calculate the solute molar mass in the JEE 2019 benzene question?

Should I use 39 g or 39.5 g to calculate molality?

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