When the nonvolatile solute #B# is added into volatile solvent #A# in a #2:5# mol ratio, the vapor pressure of #A# drops to #"250 torr"#. What will the vapor pressure of #A# above the solution be if #B# is added to #A# at a #3:5# mol ratio instead?
1 Answer
It will be
With nonvolatile solutes, particularly in ideal solutions, that's a sign that you're using Raoult's law. Furthermore, since
#P_A = chi_(A(l))P_A^"*"# where:
#chi_(A(l))# is the mol fraction of#A# in the solution phase.#P_A# is the vapor pressure of#A# above the solution.#"*"# indicates the pure substance.
You are being asked to determine the new vapor pressure due to adding more solute
Putting any solute
Thus, in light of reality, we expect that a mol ratio of
A
#chi_(A(l)) = n_A/(n_A + n_B) = 5/(5 + 2) = 0.7143#
This means that given the vapor pressure of
#P_A^"*" = P_A/(chi_(A(l)))#
#= ("250 torr")/(0.7143)#
#=# #"350 torr"# (exactly)
So, if we instead have a
#chi_(A(l)) = n_A/(n_A + n_B) = 5/(5+3) = 0.6250#
So, in this slightly different situation where we now solve for a different
#color(blue)(P_A) = chi_(A(l))P_A^"*"#
#= 0.6250cdot"350 torr"#
#=# #color(blue)("218.8 torr")#
Great, since
ALTERNATIVE METHOD
Alternatively, there is one more way to do this without solving for
If we solve implicitly for
#P_A^"*" = P_(A1)/(chi_(A(l)1))#
#P_A^"*" = P_(A2)/(chi_(A(l)2))#
where:
#P_(A1)# is the vapor pressure for#A# above the solution in instance#1# where the mol ratio is#B:A = 2:5# .#P_(A2)# is the vapor pressure for#A# above the solution in instance#2# where the mol ratio is#B:A = 3:5# .
We could then get:
#P_(A1)/(chi_(A(l)1)) = P_(A2)/(chi_(A(l)2))#
So, using the mol fractions we got above,
#P_(A2) = ((chi_(A(l)2))/(chi_(A(l)1)))P_(A1)#
#= 0.6250/(0.7143)cdot "250 torr"#
#=# #"218.8 torr"#
Just as we got before!