Question
The vapour pressure of pure water at is . When of glucose () is dissolved in of water, find (a) the mole fraction of glucose, (b) the relative lowering of vapour pressure, (c) the new vapour pressure.
Solution — Step by Step
Moles of glucose . Moles of water . Total moles .
Raoult’s law for non-volatile solute:
, relative lowering , .
Why This Works
Raoult’s law states that the partial pressure of a volatile component in a solution is its mole fraction times its pure vapour pressure. For a non-volatile solute, only the solvent contributes to the vapour, so the relative lowering equals the solute’s mole fraction.
This is a colligative property — it depends on the number of solute particles, not on what they are. A non-electrolyte gives one particle per formula unit; ionic solutes give more (Van’t Hoff factor).
Alternative Method
For dilute solutions, mole fraction can be approximated by (ignoring the solute in the denominator): . Close to — useful for quick MCQ checks.
Common Mistake
Using mass fraction instead of mole fraction in Raoult’s law. The mole fraction is what matters because the law is about particles, not mass. glucose has the same particle count as mol — using as a “fraction” is a frequent error.