Question

6. Sometimes it's hard to tell which thing is the solvent and which the solute – for example you can make mixtures of water and methanol at any proportion. Fortunately, Raoult's law doesn't care which thing you call the solvent or solute; at a given temperature, both things have vapor pressures that fit the equation    P_A = Χ_A P_A⁰ - where the right-hand side of the equation is the mole fraction times the pure liquid vapor pressure at that temperature.

a.) Calculate the vapor pressures for methanol (CH₃OH, 32.04 g/mol, P⁰ = 13.02 kPa) and water (18.02 g/mol, P⁰ = 3.20 kPa) above a mixture containing one mole each of methanol and water (in a sealed container with a head-space for the vapor).

b.) Which of the two components' (methanol's or water's) vapor pressure would be depressed the most by addition of 10 g of NaCl (58.44 g/mol) assuming complete dissolution by a mechanism similar to that which takes place in aqueous solution (similar IMFs)? Calculate the percent change in that component's vapor pressure.

Answer 1

moles of methanol = moles of water = 1

total moles = 2

mole fraction of each = 0.5

vapour pressure of solution = Pm Xm + PwXw

                                              = 13.02 x 0.5 + 3.20 x 0.5

                                              = 8.11 kPa

vapour pressure of solution    = 8.11 kPa

2)

here in methanol NaCl does not dissolve .

so NaCl i value = 1

in water will give 2 ions .

so NaCl i value = 2 .

vapour pressure is directly proportional to i value.

NaCl i value is more in water. in that vapour pressure is most depressed.