Question

calculate the composition of the vaopur above a solution containing 735 g of h20 and 245 g of CH3CH2OH at 323k. at this temperature the vapour pressure of pure water is 0.122 atm and pure ethanol is 0.292 atm

Answer 1

number of moles of water = 735/18= 40.83

number of moles of ethanol = 245/46=5.32

mole fraction of water = 40.83 /46.15= 0.88

mole fraction of ethanol = 5.32/46.15 = 0.12

According to Raoult's law

Thus the vapour pressure of solution = (0.88x 0.122 ) + (0.12x 0.292) atm

= 0.107 + 0.035

= 0.142atm

According to Dalton's law the partial pressure of a gas = its mole fraction x total pressure

Thus mole fraction of acomponent in vapor = partial pressure /total pressure

Thus

mole fraction of water vapour= 0.107/0.142=0.7535

mole fraction of ethanol in vapour = 0.035/0,142= 0.2464