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

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

Step-by-step 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)=(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