What is the mol fraction of ethylene glycol in the solution phase for an aqueous solution with a vapor pressure of #"760 torr"# if the pure vapor pressure was #"1077 torr"#?
1 Answer
I got
Raoult's law states:
#P_j = chi_j^lP_j^"*"# ,
where:
#P_j# is the partial vapor pressure above the solution coming from the component#j# in solution.#P_j^"*"# is the vapor pressure above a sample of pure liquid#j# at the same#T# as the solution.#chi_j^l# is the mol fraction of#j# in the liquid phase.
Since the total vapor pressure of the solution was given, which is lower than the pure vapor pressure of water, there was a decrease in vapor pressure.
I assume however, that the
To track that change:
#DeltaP = P_i - P_i^"*"# ,
where
Then, plugging in Raoult's law:
#DeltaP = chi_i^lP_i^"*" - P_i^"*"#
#= P_i^"*"(chi_i^l - 1)#
#= -chi_j^lP_i^"*"#
The change in pressure was:
#"760 torr"# #-# #"1077 torr" = -"317 torr"#
So, the mol fraction of ethylene glycol in the solution (not in the vapor phase) would be:
#color(blue)(chi_j^l) = -(DeltaP)/(P_i^"*")#
#= -(-"317 torr")/("1077 torr")#
#= color(blue)(0.294)#