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wariber [46]
3 years ago
6

The vapor pressure of pure water at 250C is 23.77 torr. What is the vapor pressure of water above a solution that is 1.500 m glu

cose, C6H12O6?
Chemistry
1 answer:
Olegator [25]3 years ago
7 0

Answer:

Vapor pressure of water = 23.14torr

Explanation:

When you made a solution, vapor pressure decreases following Raoult's law:

P_{solution} = X_{solvent} P_{solvent}

<em>Where P is vapor pressure and X mole fraction</em>

As vapor pressure of water is 23.77torr we must find the mole fraction of water knowing the solution is 1.500m glucose (That is 1.500 moles of glucose per kg of water = 1000g of water).

1000g of H₂O are, in moles (Molar mass: 18.02g/mol):

1000g H₂O ₓ (1mole / 18.02g) = 55.5 moles of H₂O.

As we know now the solution contains 55.5 moles of water and 1.5 moles of glucose. Thus, mole fraction of water (Solvent) is:

X_{H_2O} = \frac{55.5molesH_2O}{55.5molesH_2O + 1.5 molesGlucose} = 0.9737

Replacing in Raoult's law, pressure of water above the solution is:

P_{solution} = X_{solvent} P_{solvent}

P_{solution} = 0.9737*23.77torr

<h3>Vapor pressure of water = 23.14torr</h3>
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