The vapour pressure of 30 %(w / v) aqueous solution of glucose is __________ mm Hg at 25° C.
[Given : The density of 30 % (w/v), aqueous solution of glucose is 1.2 g cm⁻³ and vapour pressure of pure water is 24 mm Hg.]
(Molar mass of glucose is 180 g mol⁻¹.)

Solution

Chemistry ma’am · 4 steps

Use the density to find how much water is really there, then let Raoult's law finish the job.

1. Pick 100 mL. A 30% (w/v) solution has 30 g glucose in every 100 mL of solution. With density 1.2 g cm⁻³, 100 mL of solution weighs 120 g.
2. Count the moles. Water is 120 - 30 = 90 g, which is 90/18 = 5 mol. Glucose is 30/180 = 1/6 mol, about 0.167 mol.
3. Apply Raoult's law. Glucose is non-volatile, so only water contributes to the vapour pressure. That pressure is the pure-water value times the mole fraction of water.

   p = p°_(water) x_(water) = 24 × 5/(5 + 1/6) = 24 × 30/31 = 23.23 mm Hg

4. So. The vapour pressure of the solution is about 23.2 mm Hg, which rounds to 23.

The idea to keep. For a w/v solution, fix a volume, use the density to get the total mass, and subtract the solute to find the solvent. Raoult's law always needs the solvent's mole fraction.

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