Question

Calculate the mass of ethylene glycol (C₂H₆O₂) that must be added to 900 g of ethanol (C₂H₅OH) to reduce its vapor pressure by 12.0 torr at 35°C. The vapor pressure of pure ethanol at 35°C is 100 torr.

Answer 1

let M= mass of ethylene glycol,

moles od ethylene glycol= mass/molar mass, molar mass of C2H6O2= 2*12+6+32=62 g/mole

moles of ethylene glycol= M/62

mass of ethanol= 900 gm, moles of ethanol= mass/molar mass= 900/46= 19.56

from Raoults law,

vapor pressure of component in solution= mole fraction of component (C2H5OH)* pure component vapor pressure

12= x*100 x= mole fraction of component

x= 12/100=0.12

mole fraction= moles of that component/total moles

0.12= 19.56/(M/62+19.56)

0.12*(M/62+19.56)= 19.56

0.12*M/62 +0.12*19.56=19.56

0.12*M/62= 19.56*(1-0.12)=19.56*0.88

M= 19.56*0.88*62/0.12= 8893 gm