Raoult's Law and vapor pressure?
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?
We never learned this in class and the only explanation I can find doesn't make much sense. Help, it's due in 40 minutes!!!
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?
We never learned this in class and the only explanation I can find doesn't make much sense. Help, it's due in 40 minutes!!!
1 Answer
Well, for one, you should ask at least a few hours in advance...
I'm getting
Look for the numbers you are given:
- The mixture is
#77.5%"w/w"# formic acid - The pure vapor pressures of formic acid and water are
#"917 torr"# and#"974 torr"# , respectively.
You don't need anything else, other than Raoult's law:
#P_A = chi_(A(l))P_A^"*"# where:
#P_A# is the vapor pressure of component#A# above the solution (i.e. mixed into component#B# ).#P_A^"*"# is similar but is for the pure component#A# (unmixed) instead.#chi_(A(l))# is the mol fraction of component#A# in the solution phase.
The vapor pressure of just formic acid above the solution,
#P_(FA) = chi_(FA(l))P_(FA)^"*"#
The percent by mass by definition is:
#"77.5 g formic acid"/("100.00 g solution"#
#= "77.5 g formic acid"/("77.5 g formic acid" + "22.5 g water")#
From this, the mol fraction of formic acid can be gotten. Formic acid is
So:
#chi_(FA(l)) = (77.5 cancel"g formic acid" xx "1 mol"/(46.025 cancel"g"))/(77.5 cancel"g formic acid" xx "1 mol"/(46.025 cancel"g") + 22.5 cancel"g water" xx "1 mol"/(18.015 cancel"g"))#
#~~ 0.5741#
P.S. this also means that the mol fraction of water is
#chi_(w(l)) = 1 - chi_(FA(l)) = 0.4259# .
Anyways, this allows us to get the vapor pressure of formic acid above the solution:
#color(blue)(P_(FA)) = 0.5741 cdot "917 torr"#
#=# #color(blue)("526 torr")# (unrounded:#"526.45 torr"# )
What would you do to get the vapor pressure of water above the solution?
