Question 11.9: CALCULATING VAPOR-PRESSURE LOWERING How many grams of sucros...
CALCULATING VAPOR-PRESSURE LOWERING
How many grams of sucrose must be added to 320 g of water to lower the vapor pressure by 1.5 mm Hg at 25 °C? The vapor pressure of water at 25 °C is 23.8 mm Hg, and the molar mass of sucrose is 342.3 g/mol.
STRATEGY
According to Raoult’s law, P_{soln} = P_{solv} × X_{solv}, which can be rearranged to the form X_{solv} = P_{soln}/P_{solv}. This equation can then be solved to find the number of moles of sucrose and hence the number of grams.
First, calculate the vapor pressure of the solution, P_{soln}, by subtracting the amount of vapor-pressure lowering from the vapor pressure of the pure solvent, P_{solv}:
P_{soln} = 23.8 mm Hg – 1.5 mm Hg = 22.3 mm Hg
Now calculate the mole fraction of water, X_{solv}.
Since P_{soln} = P_{solv} × X_{solv}
then X_{solv} = \frac{P_{soln}}{P_{solv}} = \frac{22.3 mm Hg}{23.8 mm Hg} = 0.937
This mole fraction of water is the number of moles of water divided by the total number of moles of sucrose plus water:
X_{solv} = \frac{Moles of water}{Total moles}Since the number of moles of water is
Moles of water = 320 g × \frac{1 mol}{18.0 g} = 17.8 mol
then the total number of moles of sucrose plus water is
Total moles = \frac{moles of water}{X_{solv}} = \frac{17.8 mol}{0.937} = 19.0 mol
Subtracting the number of moles of water from the total number of moles gives the number of moles of sucrose needed:
Moles of sucrose = 19.0 mol – 17.8 mol = 1.2 mol
Converting moles into grams then gives the mass of sucrose needed:
Grams of sucrose = 1.2 mol × 342.3 \frac{g}{mol} = 4.1 × 10² g