Solutions JEE 2023: Urea Mixing and Osmotic Pressure

What is the answer to the JEE Advanced 2023 urea-mixing question?

The osmotic pressure is 682 Torr. This JEE Advanced 2023 Solutions walkthrough solves the urea-mixing numerical from Paper 2, Chemistry, chapter: Solutions. The question bank classifies it as medium difficulty, not an official exam rating.

A 50 mL portion of a 0.2 molal urea solution is mixed with 250 mL of another solution containing 0.06 g of urea. Both solutions use the same solvent. The first solution has this density at 300 K:

ρ = 1.012 g m L^{- 1}

Find the final solution’s osmotic pressure, in Torr, at 300 K. Use the supplied constants and mixing assumptions:

M_{urea} = 60 g m o l^{- 1}, R = 62 L T o r r K^{- 1} m o l^{- 1}

Δ_{m i x} H = 0, Δ_{m i x} V = 0

This is a numerical-answer question with no supplied options. Calculate the final molarity before using the osmotic-pressure formula.

Why should we start with 1000 g of solvent?

Molality counts moles of solute per kilogram of solvent, not per kilogram of solution. Choosing 1000 g of solvent gives 0.2 mol of urea immediately. This official method keeps solvent mass separate from the solution mass needed for the density calculation.

For the first solution, the chosen basis gives:

m_{solvent} = 1000 g = 1 k g

n_{urea} = 0.2 \times 1 = 0.2 m o l

Convert the urea amount into mass. Add it to the solvent mass to obtain the solution mass:

m_{urea} = 0.2 \times 60 = 12 g

m_{solution} = 1000 + 12 = 1012 g

Use the solution density to find the solution volume. Dividing solvent mass by this density would mix up two different quantities.

V_{solution} = \frac{1012 g}{1.012 g m L^{- 1}} = 1000 m L

Thus, 1000 mL of this solution contains 0.2 mol of urea. Scale that amount to the actual 50 mL portion:

n_{1} = \frac{0.2}{1000} \times 50 = 0.010 m o l

The 1000 g solvent basis is a calculation convenience, not the amount actually mixed. It establishes the urea amount per solution volume, which lets us calculate the moles in the given portion.

How do we calculate the final molarity after mixing?

Add the urea moles first, then divide by the final solution volume. Solute conservation and volume addition are separate steps. The second solution’s density is unnecessary because its urea mass and solution volume are already supplied.

Convert its urea mass into moles. Then add the amounts from both portions:

n_{2} = \frac{0.06}{60} = 0.001 m o l

n = n_{1} + n_{2} = 0.010 + 0.001 = 0.011 m o l

The question explicitly gives zero volume change on mixing. That permits us to add the two solution volumes: $Δ_{m i x} V = 0$

V = 50 + 250 = 300 m L = 0.300 L

The final molarity uses the entire mixture’s volume, not either starting portion’s volume. Divide the total urea amount by that volume in litres:

C = \frac{0.011}{0.300} = \frac{11}{300} m o l L^{- 1}

How does the osmotic-pressure calculation give 682 Torr?

Urea is a non-electrolyte, so its van’t Hoff factor is one in the supplied model. Use the final molarity and the stipulated final temperature of 300 K. No heat-balance calculation is needed.

i = 1, T = 300 K

$π = i C R T = C R T$ Substitute the concentration, gas constant and temperature. Keep the concentration as a fraction to avoid rounding:

π = \frac{11}{300} \times 62 \times 300

The numerical factors of 300 cancel. The remaining multiplication gives:

π = 11 \times 62 \times \frac{300}{300} = 11 \times 62 = 682 T o r r

The units also check. No pressure-unit conversion is needed because the supplied gas constant uses Torr:

m o l L^{- 1} \times L T o r r K^{- 1} m o l^{- 1} \times K = T o r r

Numerical entry:

682

Why does treating solution mass as solvent mass give the wrong answer?

It gives 689.44 Torr, a wrong numerical result, not a wrong option. The source question has no options. The error begins when density gives the mass of the 50 mL solution, but that mass is then treated as solvent alone.

The solution mass is:

m_{solution} = 50 \times 1.012 = 50.6 g

Using it as solvent mass produces:

n_{1, wrong} = 0.2 \times 0.0506 = 0.01012 m o l

Carrying that error through gives:

π_{wrong} = \frac{0.01012 + 0.001}{0.300} \times 62 \times 300 = 689.44 T o r r

Molality requires solvent mass; density relates solution mass to solution volume. The urea mass cannot be counted as solvent.

A different shortcut, treating 0.2 molal directly as 0.2 molar, happens to give 682 Torr here. The solvent-basis calculation proves why:

\frac{1012 g}{1.012 g m L^{- 1}} = 1000 m L

That litre contains exactly 0.2 mol of urea. Molality and molarity therefore have the same numerical value for these data, but this is not a general conversion rule. Keep the solution-mass-to-volume step: the correct answer alone does not validate the shortcut.

Which two practice questions check the same method?

Use one mass-to-molarity calculation and one dilution calculation. Both questions below are practice variations built from the supplied data, not additional verified JEE PYQs. For both, retain the same model and constants:

i = 1, M_{urea} = 60 g m o l^{- 1}, R = 62 L T o r r K^{- 1} m o l^{- 1}

Question 1: What is the osmotic pressure at 300 K of the second solution alone, containing 0.06 g urea in 250 mL of solution?

Convert the supplied solute mass into moles, then divide by solution volume in litres:

n = \frac{0.06}{60} = 0.001 m o l

C = \frac{0.001}{0.250} = 0.004 m o l L^{- 1}

π = 0.004 \times 62 \times 300 = 74.4 T o r r

Question 2: The final mixture is diluted with the same solvent to a final volume of 0.600 L at 300 K. What is its osmotic pressure?

Adding solvent leaves the urea amount unchanged: $n = 0.011 m o l$

π = \frac{0.011}{0.600} \times 62 \times 300 = 341 T o r r

The first variation checks conversion from supplied solute mass to molarity. The second checks that doubling volume at fixed moles and temperature halves osmotic pressure. Before calculating a dilution answer, write down what stays fixed: solute moles, not concentration.

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

What is the answer to the JEE Advanced 2023 urea-mixing question?

The osmotic pressure is 682 Torr, so the numerical entry is 682. The mixture contains 0.011 mol of urea in 0.300 L because the question specifies zero volume change on mixing. With urea's van't Hoff factor equal to one, osmotic pressure = (0.011/0.300) × 62 × 300 = 682 Torr.

How do I find the moles in 50 mL of the 0.2 molal urea solution?

Start with 1000 g of solvent: it contains 0.2 mol, or 12 g, of urea. The solution mass is therefore 1012 g, which occupies 1000 mL at the given density of 1.012 g/mL. The 50 mL portion contains 0.2 × 50/1000 = 0.010 mol of urea.

Why am I getting 689.44 Torr instead of 682 Torr?

That result comes from treating the 50.6 g mass of the first solution as solvent mass. Molality uses kilograms of solvent, so including urea in that mass incorrectly gives 0.01012 mol instead of 0.010 mol. Density relates solution mass to solution volume; it does not directly give solvent mass.

Can I treat 0.2 molal urea as 0.2 molar in this question?

The two concentrations happen to have the same numerical value for the supplied data, but they are not generally interchangeable. A basis of 1000 g solvent gives 0.2 mol urea and 1012 g solution, which occupies exactly 1 L at the given density. Show this conversion rather than assuming molality equals molarity.

Solutions JEE 2023: Urea Mixing and Osmotic Pressure

What is the answer to the JEE Advanced 2023 urea-mixing question?

Why should we start with 1000 g of solvent?

How do we calculate the final molarity after mixing?

How does the osmotic-pressure calculation give 682 Torr?

Why does treating solution mass as solvent mass give the wrong answer?

Which two practice questions check the same method?

Frequently asked questions

What is the answer to the JEE Advanced 2023 urea-mixing question?

How do I find the moles in 50 mL of the 0.2 molal urea solution?

Why am I getting 689.44 Torr instead of 682 Torr?

Can I treat 0.2 molal urea as 0.2 molar in this question?

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