Question
When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was x\% of vapor pressure of pure solvent. What is x\%?
JEE Main
Chemistry
Solutions
2026
JEE Main 2026 (06 April Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. 50\%
- B. 60\%
- C. 70\%
- D. 80\%
Answer
D. 80\%
Step-by-step solution
Moles of solute, n_B = 0.25 Moles of solvent, n_A = 1 Mole fraction of solvent, X_A = n_A n_A + n_B = 1 1 + 0.25 = 1 1.25 = 0.8 According to Raoult's law, the vapor pressure of the solution is given by P_s = P^0 X_A P_s = 0.8 P^0 = 80\% of P^0 Thus, the vapor pressure of the solution is 80\% of the vapor pressure of the pure solvent. Answer: 80\%
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