At room temperature, the mole fraction of a solute is 0.25 and the vapour pressure of a solvent is 0.80 atm. The lowering of vapour pressure is:

0.20

The correct answer is option 3. 0.20.

The lowering of vapor pressure (\(\Delta P\)) is given by the difference between the vapor pressure of the pure solvent (\(P^{\circ}\)) and the vapor pressure of the solution (\(P\)):

\(\Delta P = P^{\circ} - P\)

The mole fraction of the solute is 0.25 and the vapor pressure of the solvent is 0.80 atm. We can calculate the vapor pressure of the solution using Raoult's law:

\(P = x_{\text{solute}} \cdot P^{\circ}\)

where \(x_{\text{solute}}\) is the mole fraction of the solute. Since the mole fraction of the solute is 0.25, the mole fraction of the solvent is \(1 - 0.25 = 0.75\).

\(P = 0.75 \cdot P^{\circ} = 0.75 \times 0.80 \, \text{atm} = 0.60 \, \text{atm}\)

Now we can calculate the lowering of vapor pressure:

\(\Delta P = P^{\circ} - P = 0.80 \, \text{atm} - 0.60 \, \text{atm} = 0.20 \, \text{atm}\)

Therefore, the lowering of vapor pressure is 0.20 atm, which corresponds to option (3).