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?

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!