When a certain amount of solid A is dissolved in 100 g of water at 25° C to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is 23.76 mmHg. The number of moles of solute A added is _____________. (Nearest Integer)

Solution

Chemistry ma’am · 4 steps

Halving the vapour pressure means the solute's mole fraction is one-half, so its moles match water's.

1. Raoult's law. The relative lowering of vapour pressure equals the mole fraction of the solute. Use the full mole fraction here, because the lowering is far too large for the dilute approximation.

   (p° - p)/p° = n_A/(n_A + n_(water))

2. Put in numbers. With p = 1/2p°, the left side is 1/2. So n_A/(n_A + n_(water)) = 1/2, which gives n_A = n_(water). The value 23.76 mmHg cancels out and is not needed.
3. Moles of water. n_(water) = 100/18 = 5.56 mol, so n_A = 5.56 mol.
4. So. The moles of solute A added are 5.56, which rounds to 6.

The common slip: 3. Using the dilute form n_A/n_(water) = 1/2 gives n_A = 2.78, which rounds to 3, not 6. Keep n_A in the denominator.

The idea to keep. Relative lowering of vapour pressure equals the solute's mole fraction. Drop nₛₒₗᵤₜₑ from the denominator only when the lowering is small; a 50% drop is never small.

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