A solution is prepared by dissolving 60.0 g of sucrose, C12H22O11, in 250. g of water at 25°C. What is the vapor pressure of the solution if the vapor pressure of water at 25°C is 23.76 mm Hg?

23.46 mmHg is the vapor pressure for the solution

Explanation:

To solve this problem we need to apply a colligative property, which is the lowering vapor pressure.

The formula for this is: P°- P' = P° . Xm

where P' is vapor pressure for solution and P°, vapor pressure for pure solvent.

Let's determine the Xm (mole fraction for solute)

We calculate the moles of the solute and the solvent and we sum each other:

Moles of solute: 60 g /342 g/mol = 0.175 moles of sucrose

Moles of solvent: 250 g / 18 g/mol = 13.8 moles of water

Total moles: 13.8 moles + 0.175 moles = 13.975 moles

Xm for solute: 0.175 moles / 13.975 moles = 0.0125

Let's replace data in the formula: 23.76 mmHg - P' = 23.76 mmHg . 0.0125

P' = - (23.76 mmHg . 0.0125 - 23.76 mmHg) → 23.46 mmHg

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