What is the composition of a methanol (CH3OH) –propanol (C3H7OH) solution that has a vapor pressure of 174 torr at 40ºC? At 40ºC, the vapor pressures of pure methanol and pure propanol are 303 and 44.6 torr, respectively. Assume the solution is ideal. Report the composition of the liquid solution in units of mole fractions.

Explanation:

Psol = Pmeth * Xmeth + Pprop * Xprop

Where,

Psol = vapor pressure of the solution

Pmeth = vapor pressure of methanol

Pprop = vapor pressure of propanol

Xmeth = mole fraction methanol

Xprop = mole fraction propanol

Therefore,

174 = 303 Xmeth + 44.6 Xprop

Xmeth + Xprop = 1

Xmeth = 1 - Xprop

174 = 303 ( 1 - Xprop ) + 44.6 Xprop

174 = 303 - 303 Xprop + 44.6 Xprop

129 = 258.4 * Xprop

Xprop = 0.499

Xmeth = 1 - 0.499

= 0.501

Since Xprop = 0.499,

But Xprop = moles propanol / (moles propanol + moles methanol)

Since, moles propanol + moles methanol = 1,

moles of propanol, mprop = 0.499

moles methanol, mmeth = 0.501

Molar mass of propanol = (12*3) + (1*8) + 16

= 60 g/mol.

mass of propanol, Mprop = 0.499 x 60 g/mol

= 3.00 g

Molar mass of methanol = (12*1) + (1*4) + 16

= 32 g/mol.

mass of methanol, Mmeth = 0.501 x 32

= 16.1 g.

the volume increases