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

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):

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:

Replacing in Raoult's law:

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

i think it is true since you can test the answer and can use observations to complete it.