Question

at 25*C, the vapor pressure of pure water is 23.8 torr. a solution is prepared by dissolving 0.30 mol Na2SO4 in 8.0 moles of water. what is the vapor pressure of this solution at 25*C

Answer 1

Answer:

Vapor pressure of the solution is 22.9 torr at 25°C

Explanation:

When a non volatile solute is added to a pure solvent, the vapour pressure of the mixture decrease regard to the pure solvent following the equation (Raoult's law):

$P_{solution}=X_{solvent}.P^0_{solvent}$

Where vapour pressure of the pure solvent (Water) is 23.8torr at 25°C.

Mole fraction of water is the ratio between moles of water and total moles in solution. The solution has a mole fraction of water of:

$X_{water}=\frac{8.0Moles_{water}}{0.30mol_{Na_2SO_4}+8.0mol_{water}} = 0.9639$

Replacing in Raoult's law:

$P_{solution}=0.9639*23.8torr$

Vapor pressure of the solution is 22.9 torr at 25°C