80 mole percent of MgCl₂ is dissociated in aqueous solution. The vapour pressure of 1.0 molal aqueous solution of MgCl₂ at 38° C is ____________ mm Hg. (Nearest integer)
Given : Vapour pressure of water at 38° C is 50 mm Hg

Solution

Chemistry ma’am · 4 steps

First turn 80% dissociation into an i value, then use Raoult's law with 2.6 moles of solute particles.

1. Find i. MgCl₂ → Mg²⁺ + 2Cl⁻ gives n = 3 ions. With α = 0.8, i = 1 + α(n - 1) = 1 + 0.8 × 2 = 2.6.
2. Count moles. 1 molal means 1 mol of MgCl₂ in 1000 g of water. That is 1000/18 = 55.56 mol of water and 1 × 2.6 = 2.6 mol of solute particles.

   p = p° x_(water) = 50 × 55.56/(55.56 + 2.6) = 50 × 55.56/58.16

3. Put in numbers. 55.56/58.16 = 0.9553, so p = 50 × 0.9553 = 47.76 mm Hg.
4. So. The vapour pressure of the solution is about 47.76 mm Hg, which rounds to 48.

The common slip: 49. Leaving out the van't Hoff factor (taking i = 1) gives 50 × 55.56/56.56 = 49.1, which rounds to 49, not 48.

The idea to keep. Colligative properties count particles, not formula units. For a partly dissociated salt, use i = 1 + α(n-1) and multiply the solute moles by i before you apply Raoult's law.

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