The pure vapour pressure of a liquid solvent A is determined to be exactly . When an unknown, non-vo — Solutions and Colligative Properties Chemistry Question
Question
The pure vapour pressure of a liquid solvent A is determined to be exactly $0.80\text{ atm}$. When an unknown, non-volatile solid substance B is dissolved into the solvent, the equilibrium vapour pressure of the resulting dilute solution drops to $0.60\text{ atm}$. What is the theoretical mole fraction of component B in the solution? (Multiply your final calculated answer by 100 to report it as a simple integer percentage).
💡 Solution & Explanation
For a solution containing a non-volatile solute, Raoult's law states that the relative lowering of vapour pressure (RLVP) is exactly equal to the mole fraction of the solute ($X_B$). $X_B = \frac{P_A^0 - P_s}{P_A^0} = \frac{0.80 - 0.60}{0.80} = \frac{0.20}{0.80} = 0.25$. Converting this fraction to a percentage: $0.25 \times 100 = 25$.