This is Raoult law & vapor pressure problem: high school

2.7 moles of Tang are added to a pitcher containing 3.28 liters of H2O at a temperature of 25°C. The vapor pressure of water alone is 23.8 mm Hg at 25o C. What is the new vapor pressure of the Tang solution?

As per Rault's law, when a solute is dissolved, the vapor pressure of the solution is equal to the product of the mole fraction of the solvent and the vapor pressure of the pure solvent.

[tex]P_{soluiton}=X_{solvent}\times P^0[/tex]

1. Calculate the mole fraction of solvent

[tex]X_{solvent}=\dfrac{\text{moles of solvent}}{\text{moles of solvent + moles of solute}}[/tex]

a) Moles fo solvent

The solven is water

3.28 liters of H₂O = 3,280 g of H₂O (taking density equal to 1.000 g/mol.

number of moles = mass in grams/molar mass

molar mass of H₂O = 18.015 g/mol

number of moles of H₂O = 3,280g / (18.015g/mol) = 182.07 mol

b) Moles of solute

The solute is tang:

2.7 mol (given)

c) Mole fraction of solvent:

[tex]X_{solvent}=\dfrac{182.07}{182.07+2.7}\approx0.985387[/tex]

2. Vapor pressure of the solution

[tex]P_{solution}=0.985387\times 23.8mmHg\approx23.5mmHg[/tex]

This is Raoult law & vapor pressure problem: high school 2.7 moles of Tang are added to a pitcher containing 3.28 liters of H2O at a temperature of 25°C. The vapor pressure of water alone is 23.8 mm Hg at 25o C. What is the new vapor pressure of the Tang solution?

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This is Raoult law & vapor pressure problem: high school

2.7 moles of Tang are added to a pitcher containing 3.28 liters of H2O at a temperature of 25°C. The vapor pressure of water alone is 23.8 mm Hg at 25o C. What is the new vapor pressure of the Tang solution?