A set of solutions is prepared using 180 g of water as a solvent and 10 g of different nonvolatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order :
[Given, molar mass of A = 100 g mol –1 ; B = 200 g mol –1 ; C = 10,000 g mol –1 ]

Solution

Chemistry ma’am · 3 steps

Same 10 g of each solute, so the lightest solute gives the most moles and the biggest lowering.

1. Raoult's law. For a nonvolatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute. It depends only on how many solute particles there are, not on what they are.

   (p° - p)/p° = xₛₒₗᵤₜₑ = n/(n + N)

2. Moles of water. N = 180/18 = 10 mol in every solution. Then n = 10/M for each solute.

   Solute	Molar mass	Moles, n = 10/M	Relative lowering, n/(n + 10)
   A	100 g mol⁻¹	0.1	0.1/10.1 ≈ 9.9 × 10⁻³, largest
   B	200 g mol⁻¹	0.05	0.05/10.05 ≈ 5.0 × 10⁻³
   C	10 000 g mol⁻¹	0.001	0.001/10.001 ≈ 1.0 × 10⁻⁴, smallest

3. So. With equal masses, more moles means a lower molar mass. The relative lowering follows A > B > C, which is option (c).

The idea to keep. Relative lowering of vapour pressure counts particles. When the solute masses are equal, the solute with the smaller molar mass gives more moles and so lowers the vapour pressure more.

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