Question

Methanol and ethanol form a nearly ideal solution at 300 K. A solution is made by mixing 32 g of methanol and 23 g of ethanol at 300 K. Calculate the partial pressures of its constituents and the total pressure of the solution (At 300 K, \[\ce{p^\circ_{CH_3OH}}\] = 90 mm Hg, \[\ce{p^\circ_{C_2H_5OH}}\] = 51 mm Hg)

Answer 1

Mass of methanol = 32 g

Mass of ethanol = 23 g

Temperature = 300 K

Vapour pressure of pure methanol = 90 mm Hg 

Vapour pressure of pure ethanol = 51 mm Hg

Molar mass of methanol = 32.04 g/mol

Molar mass of ethanol = 46.07 g/mol

Moles of methanol = `32/32.04`

= 0.9987

Moles of ethanol = `23/46.07`

= 0.4992

Total moles = 0.9987 + 0.4992

= 1.4979 mol

Mole fraction of methanol = `09987/1.4979`

= 0.6667

Mole fraction of ethanol = `0.4992/1.4979`

= 0.3333

Partial pressure of methanol = 0.6667 × 90

= 60.003 mm Hg

Partial pressure of ethanol = 0.3333 × 51

= 16.998 mm Hg

Total pressure = 60.003 + 16.998

= 77.001 mm Hg