In: Chemistry
1) At a pressure of 760 torr, formic acid (HCOOH; boiling point = 100.7 oC) and water (H2O; boiling point = 100.0 oC) form an azeotropic mixture, boiling at 107.1 oC, that is 77.5% by mass formic acid. At the boiling point of the azeotrope (107.1 oC), the vapor pressure of pure formic acid is 917 torr, and that of pure water 974 torr. If the solution obeyed Raoult's law for both components, what would be the vapor pressure (in torr) of formic acid at 107.1 oC?
2) If the solution obeyed Raoult's law for both components, what would be the vapor pressure of the water (in torr) at 107.1 oC?
3) If the solution obeyed Raoult's law for both components, what would be the total vapor pressure (in torr) at 107.1 oC?
1) let the mixture be 100 g
then
mass of HCOOH = 77.5
mass of water = 22.5
now
we know that
moles = mass / molar mass
so
moles of HCOOH = 77.5 / 46 = 1.6848
moles of H20 = 22.5 / 18 = 1.25
now
total moles = 1.6848 + 1.25 = 2.9348
now
mole fraction of HCOOH = moles of HCOOH / total moles
mole fraction of HCOOH = 1.6848 / 2.9348
mole fraction of HCOOH = 0.574
now
we know that
vapor pressure of HCOOH = mole fraction x vapor pressure of pure HCOOH
so
vapor pressure of HCOOH = 0.574 x 917 = 526.3
so
vapor pessure of HCOOH is 526.3 torr
2)
now
mole fraction of HCOOH + mole fraction of H20 = 1
so
0.574 + mole fraction of H20 =1
mole fraction of H20 = 0.426
now
vapor pressure of H20 = mole fraction x vapor pressure of pure H20
vapor pressure of H20 = 0.426 x 974
vapor pressure of H20 = 414.924
so
vapor pressure of H20 is 414.924 torr
3) now
total vapor presure = vapor pressure of HCOOH + vapor pressure of H20
= 526.3 + 414.924
= 941.2
so
the total vapor presure is 941.2 torr