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(A) According to Raoult's Law,the relative lowering of vapor pressure is equal to the mole fraction of the solute,but the ratio of the vapor pressure

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$2/3$

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(A) According to Raoult's Law,the relative lowering of vapor pressure is equal to the mole fraction of the solute,but the ratio of the vapor pressure of the solution $(P_{solution})$ to the vapor pressure of the pure solvent $(P^0_{solvent})$ is equal to the mole fraction of the solvent $(X_{solvent})$.$X_{solvent} = \frac{n_{solvent}}{n_{solvent} + n_{solute}}$Given: $n_{solvent} = 2 \ mol$,$n_{solute} = 1 \ mol$$\frac{P_{solution}}{P^0_{solvent}} = \frac{2}{2 + 1} = \frac{2}{3}$

One mole of a non-volatile solute is dissolved in two moles of water. What is the vapor pressure of this solution relative to that of pure water?

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