Question

Calculate the mass of ethylene glycol (C₂H₆O₂ - molar mass =62.07 g/mol) that must be added to 1.00 kg of ethanol (C₂H₅OH- molar mass =46.07 g/mol) to reduce its vapor pressure by 10.0 torr at 35 degree C. The vapor pressure of pur ethanol at 35 degree C is 100 torr.

Answer 1

Mass of ethanol =1 kg =1000gms =1000/46.07 gmoles = 21.71gmoles

vapor pressure of ethanol due to addition of glycol = 100-10 =90 torr

From Raoult's law

vapor pressure in solution =pure component vapor pressure * mole fraction of ethanol

90 =100* mole fraction of ethanol

mole fraction ethanol = 0.9 =moles of ethanol/ ( moles of ethylene glycol+ moles of ethanol)

0.9 =21.71/( 21.71+moles of ethylene glycol)

21.71+ mole of etthylene glycol = 21.71/0.9=24.12

moles of ethylene glycol = 24.12-21.71=2.41 gmoles ,it mass = moles * molecular weight = 2.41*62.07=149.58gms