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
Jun 1, 2017

It will be #"218.8 torr"# (or #"mm Hg"#).


With nonvolatile solutes, particularly in ideal solutions, that's a sign that you're using Raoult's law. Furthermore, since #B# is nonvolatile, the vapor pressure is purely due to #A# in the vapor phase---no #B# is in the vapor phase.

#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 #B# into solvent #A#, given a vapor pressure in an initial state.

Putting any solute #B# into a solvent lowers the solvent's resultant vapor pressure, #P_A#, relative to the solvent's original vapor pressure, #P_A^"*"#. A more detailed explanation can be found here.

Thus, in light of reality, we expect that a mol ratio of #3:5# instead of #2:5# means that #P_(A)# becomes lower than #"250 mm Hg"# (or #"torr"#). If it's not, we got an error!

A #2:5#, #B:A# molar ratio implies a mol fraction of #A# in the solution phase of:

#chi_(A(l)) = n_A/(n_A + n_B) = 5/(5 + 2) = 0.7143#

This means that given the vapor pressure of #"250 torr"# for #P_A#, we can get #P_A^"*"#, the pure vapor pressure of #A# (by itself):

#P_A^"*" = P_A/(chi_(A(l)))#

#= ("250 torr")/(0.7143)#

#=# #"350 torr"# (exactly)

So, if we instead have a #3:5#, #B:A# ratio, we have a different mol fraction of #A# in the solution phase:

#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 #P_A# (while #P_A^"*"# remains the same at the same temperature), we get:

#color(blue)(P_A) = chi_(A(l))P_A^"*"#

#= 0.6250cdot"350 torr"#

#=# #color(blue)("218.8 torr")#

Great, since #"218.8 torr" < "250 torr"#, this makes physical sense!

ALTERNATIVE METHOD

Alternatively, there is one more way to do this without solving for #P_A^"*"#. You can treat it similarly to an ideal gas law problem, where you could write, for example, #P_1V_1 = P_2V_2#, or #V_2/T_2 = V_1/V_1#.

If we solve implicitly for #P_A^"*"#, which does not change under the same surrounding conditions, we get:

#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!