Question

What is the vapor pressure of the solution that contains 25g of mannose (non-volatile non-electrolyte MW 180.16 g/mol) and 75g of water (MW 18)? The vapor pressure of pure water is 23.8 torr.

I know the answer is 23, but I do not know how to get it.

Answer 1

moles of mannose = mass / molar mass of it

             = 25 /180.16 = 0.1387655

moles of H2O = mass of H2O / molar mas sof H2O = 75/18 = 4.167

mol fraction of mannose= moles of mannose / ( total moles present in solution)

             = 0.1387655 / ( 0.1387655+4.167)

            = 0.03223

we have formula Po-Ps = Po x ( Xb)

where Po = vapor pressure of pure olvent = 23.8 torr

Ps = vapor pressure of solution

Xb = mol fraction of solute = 0.03223

we find Ps now

23.8-Ps = 0.03223 x 23.8

Ps = 23 torr