The theoretical equilibrium vapour pressure of a pure liquid solvent A is . When an unknown, strictl — Solutions and Colligative Properties Chemistry Question
Question
The theoretical equilibrium vapour pressure of a pure liquid solvent A is $0.80\text{ atm}$. When an unknown, strictly non-volatile solute B is added to this solvent, the equilibrium vapour pressure drops to exactly $0.60\text{ atm}$. What is the exact mathematical mole fraction of the solute B ($\chi_B$) in the resulting solution?
Answer: C
💡 Solution & Explanation
For a non-volatile solute, Raoult's law relates the relative lowering of vapour pressure directly to the mole fraction of the solute: $\frac{P_A^0 - P_s}{P_A^0} = \chi_B$. Substituting the given values: $\chi_B = \frac{0.80 - 0.60}{0.80} = \frac{0.20}{0.80} = \frac{1}{4} = 0.25$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes