Question

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?

Answer 1

• 23.5 mmHg

Step-by-step explanation:

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.

`P_(soluiton)=X_(solvent)* P^0`

1. Calculate the mole fraction of solvent

`X_(solvent)=\frac{\text{moles of solvent}}{\text{moles of solvent + moles of solute}}`

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:

`X_(solvent)=(182.07)/(182.07+2.7)\approx0.985387`

2. Vapor pressure of the solution

`P_(solution)=0.985387* 23.8mmHg\approx23.5mmHg`