Above every liquid sits a vapour, and three laws tell you how much.

Raoult's law, $P = x\,P^{0}$ — P equals x P-nought. $P$ is the vapour pressure of the solvent above the solution, $P^{0}$ (say “P-nought”) is the vapour pressure of the pure solvent at that temperature, and $x$ is the mole fraction of the SOLVENT — not the solute. That is the structural trap of this lesson, and it is worth saying twice: Raoult reads the solvent's fraction, because the solvent is what evaporates. Dissolve something non-volatile and $x < 1$, so $P$ always falls below $P^{0}$. That drop is why the boiling point had to climb in the last lesson.

Henry's law, $C = HP$, runs the other direction: a gas dissolving INTO a liquid. $C$ is the dissolved concentration in mol/m³, $P$ is that gas's partial pressure above the liquid in pascals, and $H$ is the Henry solubility constant in mol/(m³·Pa) — 1.3 × 10⁻⁵ for oxygen in water at 25 °C, 3.3 × 10⁻⁴ for carbon dioxide, which is precisely why a bottle fizzes when you release the pressure above it. Note the pascals: $H$ is quoted per Pa, so kilopascals must be converted before they go in.

The curve is Clausius–Clapeyron, two-point form: $\ln\!\left(\dfrac{P_2}{P_1}\right) = -\dfrac{\Delta H_{vap}}{R}\left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right)$. $P_1$ and $P_2$ are the pure liquid's vapour pressures at absolute temperatures $T_1$ and $T_2$ — subscripts back to naming two STATES of one liquid — and $\Delta H_{vap}$ is its molar enthalpy of vaporisation in J/mol, with $R = 8.314\ \mathrm{J/(mol \cdot K)}$. Vapour pressure climbs exponentially with temperature, which is why the plot that straightens it is $\ln P$ against $1/T$, and why water boils at 71 °C on top of Everest. Two units to guard: $\Delta H$ in joules to match R, and every T in kelvin, because a reciprocal temperature is meaningless from any other zero.