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14 July, 11:05

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?

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  1. 14 July, 11:18
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    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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