HardChemistryHigh School (11–12)Solutions › Solubility, Vapour Pressure and Raoult's LawNumerical

12.0 g of a non-volatile compound is dissolved in 90.0 g of water at 298 K…

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$12.0\ \text{g}$ of a non-volatile compound is dissolved in $90.0\ \text{g}$ of water at $298\ \text{K}$ . The vapour pressure falls from $3.17\ \text{kPa}$ for pure water to $3.12\ \text{kPa}$ for the solution. Find the moles of water, the moles of the compound and its molar mass.

Need a reminder? The key ideaShow

For two volatile liquids, Raoult's law gives each partial pressure as $p = x\,p^\circ$ , and the total pressure is the sum, $P = p_A + p_B$ . The vapour has $y_A = \dfrac{p_A}{P}$ . For a non-volatile solute, $p = x_{solvent}\,p^\circ$ and $\dfrac{p^\circ - p}{p} = \dfrac{n_B}{n_A}$ . Mole fractions use moles, so divide each mass by its molar mass first.

The worked solution

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← 18.4 g of glycerol, with molar mass 92.0 g/mol, is dissolved in 90.0 g of water at 298 K…Liquids A and B have vapour pressures of 35.0 kPa and 17.0 kPa when pure, at one temperature… →

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