Molality of an aqueous solution of urea is 4.44 m. Mole fraction of urea in solution is x × 10⁻³, Value of x is ________. (Integer answer)

Solution

Chemistry ma’am · 3 steps

Take exactly 1 kg of water: molality then tells you the moles of urea directly.

1. Pick 1 kg solvent. Molality is moles of solute per kilogram of solvent. So in 1000 g of water there are 4.44 mol of urea.
2. Moles of water. n_(H₂O) = 1000/18 = 55.56 mol. Total moles in the solution = 4.44 + 55.56 = 60.0 mol.

   xᵤᵣₑₐ = nᵤᵣₑₐ/(nᵤᵣₑₐ + n_(H₂O)) = 4.44/(4.44 + 55.56) = 4.44/60.0 = 0.074

3. So. xᵤᵣₑₐ = 74 × 10⁻³, so x = 74.

The common slip: 80. Dividing the moles of urea by the moles of water alone (4.44/55.56) gives about 0.080, so 80. The denominator must be the total moles, urea plus water.

The idea to keep. To go from molality to mole fraction, assume 1 kg of solvent. Moles of solute equal the molality, and moles of water are 1000/18 = 55.56.

Answer from the NTA final answer key. Solution by PrepWiser.