# 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?

Jun 1, 2017

It will be $\text{218.8 torr}$ (or $\text{mm Hg}$).

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.

${P}_{A} = {\chi}_{A \left(l\right)} {P}_{A}^{\text{*}}$

where:

• ${\chi}_{A \left(l\right)}$ is the mol fraction of $A$ in the solution phase.
• ${P}_{A}$ is the vapor pressure of $A$ above the solution.
• $\text{*}$ 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, ${P}_{A}^{\text{*}}$. 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 $\text{250 mm Hg}$ (or $\text{torr}$). 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:

${\chi}_{A \left(l\right)} = {n}_{A} / \left({n}_{A} + {n}_{B}\right) = \frac{5}{5 + 2} = 0.7143$

This means that given the vapor pressure of $\text{250 torr}$ for ${P}_{A}$, we can get ${P}_{A}^{\text{*}}$, the pure vapor pressure of $A$ (by itself):

${P}_{A}^{\text{*}} = {P}_{A} / \left({\chi}_{A \left(l\right)}\right)$

$= \frac{\text{250 torr}}{0.7143}$

$=$ $\text{350 torr}$ (exactly)

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

${\chi}_{A \left(l\right)} = {n}_{A} / \left({n}_{A} + {n}_{B}\right) = \frac{5}{5 + 3} = 0.6250$

So, in this slightly different situation where we now solve for a different ${P}_{A}$ (while ${P}_{A}^{\text{*}}$ remains the same at the same temperature), we get:

$\textcolor{b l u e}{{P}_{A}} = {\chi}_{A \left(l\right)} {P}_{A}^{\text{*}}$

$= 0.6250 \cdot \text{350 torr}$

$=$ $\textcolor{b l u e}{\text{218.8 torr}}$

Great, since $\text{218.8 torr" < "250 torr}$, this makes physical sense!

ALTERNATIVE METHOD

Alternatively, there is one more way to do this without solving for ${P}_{A}^{\text{*}}$. You can treat it similarly to an ideal gas law problem, where you could write, for example, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, or ${V}_{2} / {T}_{2} = {V}_{1} / {V}_{1}$.

If we solve implicitly for ${P}_{A}^{\text{*}}$, which does not change under the same surrounding conditions, we get:

${P}_{A}^{\text{*}} = {P}_{A 1} / \left({\chi}_{A \left(l\right) 1}\right)$

${P}_{A}^{\text{*}} = {P}_{A 2} / \left({\chi}_{A \left(l\right) 2}\right)$

where:

• ${P}_{A 1}$ is the vapor pressure for $A$ above the solution in instance $1$ where the mol ratio is $B : A = 2 : 5$.
• ${P}_{A 2}$ 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:

${P}_{A 1} / \left({\chi}_{A \left(l\right) 1}\right) = {P}_{A 2} / \left({\chi}_{A \left(l\right) 2}\right)$

So, using the mol fractions we got above,

${P}_{A 2} = \left(\frac{{\chi}_{A \left(l\right) 2}}{{\chi}_{A \left(l\right) 1}}\right) {P}_{A 1}$

$= \frac{0.6250}{0.7143} \cdot \text{250 torr}$

$=$ $\text{218.8 torr}$

Just as we got before!