Vapour Pressure of Binary Solutions of Solids in Liquids

Now consider the other important binary case: a nonvolatile solute dissolved in a volatile liquid solvent. Because the solute itself contributes essentially no vapour, the vapour pressure of the resulting solution comes only from the solvent, and adding the nonvolatile solute lowers the solvent's vapour pressure -- this is exactly why, for example, the vapour pressure of a salt solution is lower than that of pure water.

Since only the solvent evaporates, the solution's vapour pressure is proportional to the solvent's own mole fraction $x_A$:

At $x_A=1$ (pure solvent), $P_{solution}$ must equal the pure solvent's vapour pressure $P^\circ_{solvent}$, so $k=P^\circ_{solvent}$:

Rearranging:

Subtracting both sides from 1:

Since $x_A+x_B=1$ (so $1-x_A=x_B$), and the left side simplifies to a single fraction, this becomes

The left side of this equation, $\dfrac{P^\circ_{solvent}-P_{solution}}{P^\circ_{solvent}}$, is called the relative lowering of vapour pressure -- and equation 9.16 shows it is numerically equal to the solute's mole fraction, $x_B$. This lets Raoult's law be restated in an equivalent, purely solute-focused form: "the relative lowering of vapour pressure of an ideal solution containing a nonvolatile solute is equal to the mole fraction of the solute, at a given temperature." This relative lowering is developed further as a colligative property in section 9.9.1, including how it is used to determine an unknown solute's molar mass.

Comparing Raoult's law and Henry's law. For a solution of a nonvolatile solute, Raoult's law reads

while Henry's law (section 9.5) reads

$p_{solute} = K_H\,x_{solute\ in\ solution} \qquad (9.18)$ …