JEE Main 2021 · 17 March · morning · Numerical answer · +4 / −0
The mole fraction of a solute in a 100 molal aqueous solution is ___________ × 10⁻². (Round off to the Nearest Integer).
[Given : Atomic masses : H : 1.0 u, O : 16.0 u ]
Answer and solution
Answer 64
Solution
Chemistry ma’am · 3 steps
Molality means per kilogram of water, so take 1 kg of water and count the moles on both sides.
- Read molality. A 100 molal solution has 100 mol of solute dissolved in every 1 kg of water. So take exactly 1000 g of water as your sample.
- Moles of water. The molar mass of H₂O is 2 × 1 + 16 = 18 g mol⁻¹. So n_(H₂O) = 1000/18 = 55.56 mol.
xₛₒₗᵤₜₑ = nₛₒₗᵤₜₑ/(nₛₒₗᵤₜₑ + n_(H₂O)) = 100/(100 + 55.56) = 100/155.56 = 0.643
- So. xₛₒₗᵤₜₑ = 0.643 = 64.3 × 10⁻², which rounds to 64.
The common slip: 36. Calculating the mole fraction of water, 55.56/155.56 = 0.357, gives 36, not 64. The question asks for the solute.
The idea to keep. To turn molality into mole fraction, assume 1 kg of solvent. Then moles of solute equal the molality, and moles of solvent are 1000 divided by its molar mass.
Answer from the NTA final answer key. Solution by PrepWiser.