Question

the vapore pressure of ethanel ( C2 H5 OH) at 20 degree C is 44mmHg and the vaper pressure of methol ( CH3 OH) at hte same tempature is 94mmHg. a mixture of 26.9g of methanol and 49.7g of ethanol is prepared and can be asumed to behave as an ideal solution.

calculate the vapor pressure of methanol and ethanol above this solution at 20 degree C?

calculate the mole fraction of methanol and ethanol in the vapor above this solution at 20 degree .C?

Answer 1

Given mass of methanol,CH₃OH is = 26.9 g

          mass of ethanol,C₂H₅OH is = 49.7 g

Molarmass of methanol,CH₃OH is = 12 + (3x1) + 16 + 1 = 32 g/mol

Molarmass of ethanol,C₂H₅OH is = (2x12) + (5x1) + 16 + 1 = 46 g/mol

Number of moles of methanol , n = mass/molarmass

                                                    = 26.9 g / 32 (g/mol)

                                                   = 0.84 mol

Number of moles of ethanol , n' = mass/molarmass

                                                 = 49.7 g / 46 (g/mol)

                                                 = 1.08 mol

Total number of moles , N = n + n'

                                        = 0.84 + 1.08

                                        = 1.92 moles

Mole fraction of methanol , X = n / N

                                             = 0.84 / 1.92

                                           = 0.44

Mole fraction of ethanol , X' = n' / N

                                             = 1.08 / 1.92

                                           = 0.56

According to Raoult's law , relative lowering of vapor pressure = mole fraction of solute.

                                                                        (P_o - P) / P_o   = X

Since the number of moles of methanol is less than that of ethanol so the solute is methanol.

Where

P_o = vapour pressure of pure ethanol = 44 mm Hg

p = vapor pressure of the solution = ?

X = mole fraction of methanol = 0.44

Plug the values we get    (44- P) / 44 = 0.44

                                        44 - P = 19.36

                                                         P = 44 - 19.36

                                                             = 24.64 mm Hg

So the vapor pressure of solution is 24.64 mm Hg