When a specific amount of propanol (C3H8O) is added to 110.0 g of pure water at 60°C, the vapor pressure of water over the solution is lowered by 1.596 kPa. Given the vapor pressure of water at 60°C is 19.932 kPa, what is the mass of propanol added?

Answer 1

Psoln = Xsolvent*Po pure solvent.
You have Po H2O. Psoln = Po-1.596. Substitute and solve for Xsolvent.

Then XH2O = [nH2O/(nH2O + nPropanol)]
Substitute and solve for nPropanol and convert that to grams. Post your work if you gst stuck.