Chemistry: When 1 mol of a nonvolatile, nondissociating solute is dissolved in 3 mol of volatile solvent, the ratio of vapor pressure of the solution to that of the pure solvent (at the same temperature) is approximately:

When 1 mol of a nonvolatile, nondissociating solute

ayoismeisalex

30.06.2021

$\frac{P_{solution}}{P_{solvent}^{vap}} =0.75$

Explanation:

Hello there!

In this case, since the solvation of a nonvolatile-nondissociating solute in a volatile solvent is modelled via the Raoult's law:

$P_{solution}=x_{solvent}P_{solvent}^{vap}$

Thus, we can calculate the ratio of the vapor pressure of the solution to that of the pure solvent, mole fraction, as shown below:

$x_{solvent}=\frac{P_{solution}}{P_{solvent}^{vap}} =\frac{n_{solvent}}{n_{solute}+n_{solvent}}$

Thus, we plug in the moles of solvent and solute to obtain:

$\frac{P_{solution}}{P_{solvent}^{vap}} =\frac{3}{3+1}\\\\ \frac{P_{solution}}{P_{solvent}^{vap}} =0.75$

Regards!

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