# 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!