One mole of sugar is dissolved in three moles of water at 298 K. The relative lowering of vapour pressure is:

Question

Accepted Answer

According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute in the solution .

Relative lowering of vapour pressure = $\frac{p_1^0 - p_1}{p_1^0} = x_2$
Where $x_2$ is the mole fraction of the solute (sugar).

Given:
Moles of solute (sugar), $n_2 = 1 	ext{ mol}$
Moles of solvent (water), $n_1 = 3 	ext{ mol}$

Mole fraction of solute, $x_2 = \frac{n_2}{n_1 + n_2} = \frac{1}{1 + 3} = \frac{1}{4} = 0.25$

Therefore, the relative lowering of vapour pressure is 0.25.