What is the number of ions produced per formula unit of the salt in the JEE Advanced 2022 relative lowering of vapour pressure numerical?
0.1 mol of an ionic salt dissolved in 1.8 kg water at 35 °C is 90 % dissociated. The vapour pressure of the resulting solution is 59.724 mm Hg. The vapour pressure of pure water at the same temperature is 60.000 mm Hg. Find the number of ions produced per formula unit of the salt.
The official solution gives exactly 5. Students familiar with Raoult's law still err when writing the expression for effective moles at α = 0.9.
How does the official solution reach exactly 5 for this relative lowering numerical?
The official solution reaches N = 5 by calculating relative lowering as 0.0046, solvent moles as 100 mol, i = 0.10 + 0.90N, effective moles as 0.01 + 0.09N, and solving the linear equation after the dilute approximation.
Moles of water = 1800 g / 18 g mol⁻¹ = 100 mol.
Let N be the number of ions produced per formula unit. With α = 0.90 the van't Hoff factor is i = 1 + α(N - 1) = 0.10 + 0.90N.
Effective moles of solute particles = 0.1 × (0.10 + 0.90N) = 0.01 + 0.09N.
Using the dilute approximation:
This gives 0.46 = 0.01 + 0.09N → 0.45 = 0.09N → N = 5.
This sequence matches the verified official worked solution line for line. Every vapour-pressure digit is kept until the final subtraction.
Why does writing total particles as αN produce a wrong numerical answer near 5?
Writing total particles as αN instead of 1 + α(N - 1) replaces i = 0.10 + 0.90N with i = 0.9N. Effective moles then become 0.09N.
The equation changes to:
This yields N ≈ 5.11 which cannot be the integer answer expected. The mistake ignores the undissociated solute molecules that still contribute. Always begin with the full expression 1 + α(N - 1) before substituting α = 0.9.
Why does the dilute-solution approximation work for this particular data set?
The dilute-solution approximation holds because effective solute moles at N = 5 is 0.46 mol while solvent moles = 100 mol. This makes n_solute / (n_solute + n_water) ≈ n_solute / n_water with error from the approximation less than 0.5 %.
JEE Advanced expects this simplification for aqueous solutions at this concentration. The data set was chosen so the approximation introduces negligible error and avoids an unnecessary quadratic.
What similar colligative properties numericals should you solve next to test the same skills?
Solve these three immediately after the 2022 vapour-pressure numerical. They reuse the identical van't Hoff setup.
- A 1:1 electrolyte shows α = 0.8 in a 0.05 molal aqueous solution with Kf = 1.86 K kg mol⁻¹. Calculate the freezing-point depression. Write i = 1 + 0.8(2 - 1) first, then multiply by Kf and molality.
- Find the osmotic pressure at 27 °C produced by 0.01 mol of a salt that gives 3 ions per formula unit at 80 % dissociation dissolved in 1 L of solution. Use i = 1 + 0.8(3 - 1) = 2.6 and the relation π = iCRT with R = 0.0821 L atm mol⁻¹ K⁻¹.
- At identical molality and complete dissociation, compare relative lowering of vapour pressure for one salt producing 2 ions versus another producing 4 ions. The ratio follows directly from the ratio of their i values.
Work each one with pen and paper before checking the result. The algebra mirrors the 2022 steps exactly.
You can search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free) to locate more examples from the solutions chapter.
How should I prepare relative lowering of vapour pressure with partial dissociation for JEE Advanced?
Always write i = 1 + α(N - 1) before substituting α = 0.9. This habit prevents the particle-count error that produces 5.11 instead of 5. Calculate solvent moles first: 1.8 kg water is exactly 100 mol.
Keep three-decimal vapour-pressure data as is until the final subtraction. Early rounding turns 0.0046 into 0.005 and changes the answer. Practise at least five past-year numericals that mix Raoult's law with van't Hoff factor so the 150-second calculation becomes routine.
Focus on the linear equation after the approximation. Drill the sequence on mixed electrolyte problems until the integer appears without hesitation.
Which exact formulas from this numerical must you keep ready for revision?
- Relative lowering = (P⁰ - P)/P⁰ = x_solute (effective)
- For dilute solution x_solute ≈ n_effective / n_water
- van't Hoff factor i = 1 + α(N - 1)
- Effective moles = (moles of salt) × i
Run through this list before every mock test that includes the solutions chapter. Apply these four lines in order without deviation.
Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).
Read next: Reflection and Refraction JEE 2019: Concave Mirror in Water.
Frequently asked questions
What is the number of ions produced per formula unit in the JEE Advanced 2022 vapour pressure numerical?
The official solution gives exactly 5. It is obtained by using the van't Hoff factor i = 1 + α(N-1) with α=0.9. This leads to the equation 0.0046 = (0.01 + 0.09N)/100 which solves to N=5.
Why does writing total particles as αN give wrong answer near 5 in solutions JEE 2022?
Writing total particles as αN replaces i with 0.9N instead of 0.1 + 0.9N. This ignores undissociated solute molecules that still contribute to the particle count. The resulting equation yields N ≈ 5.11 instead of the expected integer 5.
Why does the dilute-solution approximation work for the JEE 2022 vapour pressure data?
Effective solute moles at N=5 is 0.46 mol while solvent moles equal 100 mol. This makes the approximation n_solute/n_water sufficiently accurate with error below 0.5 %. JEE Advanced selects such data to avoid solving a quadratic equation.
What is the correct van't Hoff factor expression to use in solutions JEE 2022 numericals?
Always begin with i = 1 + α(N - 1). For α = 0.9 this simplifies to i = 0.1 + 0.9N. This prevents the common particle-count error and leads directly to the official answer of 5 without hesitation.