When 0.25 mole of a non-volatile, non-ionizable solute is dissolved in 1 mole of a solvent, the vapour pressure of the solution becomes x% of the vapour pressure of the pure solvent.

Find the value of x.

According to Raoult's Law:

P = X_solvent P°

where:

• P = vapour pressure of solution
• P° = vapour pressure of pure solvent
• X_solvent = mole fraction of solvent

Moles of solvent:

= 1

Moles of solute:

= 0.25

Total moles:

= 1 + 0.25

= 1.25

Mole fraction of solvent:

X_solvent = 1/1.25

= 0.8

Therefore:

P/P° = 0.8

= 80%

Final Answer

Option (2) 80%