Put trays in the absorber instead of packing and the question changes from how many metres to how many stages. Kremser answered it in 1930, and his equation replaced stepping off a diagram by hand:

$N = \dfrac{\ln\left[\dfrac{y_1}{y_2}\left(1 - \dfrac{1}{A}\right) + \dfrac{1}{A}\right]}{\ln A}$ — N equals the log of the bracket, over log A. $N$ is the theoretical stage count, $A$ the absorption factor from the last lesson, $y_1$ the solute mole fraction in the gas entering at the bottom and $y_2$ the fraction leaving at the top. Only the ratio $y_1/y_2$ matters, which is to say only the recovery matters.

Set it beside Colburn's transfer-unit count and the comparison teaches more than either equation alone. The numerator is identical. The denominator is $\ln A$ for stages and $1 - 1/A$ for transfer units, and that single difference is the whole story: a transfer unit and an equilibrium stage are genuinely different objects, not two scales for one thing. Watch them behave differently at the extremes. Throw unlimited solvent at the column and the stage count falls toward zero, while $N_{OG}$ never goes below $\ln(y_1/y_2)$ — because the gas still has to be diluted by that factor, and each transfer unit only ever buys one e-fold of it. At the other end, $A = 1$ makes both expressions indeterminate, and both have the same limit: $y_1/y_2 - 1$.

Two assumptions ride inside the equation, and both are worth stating out loud. The equilibrium line is taken as straight across the whole range, which is fair for a dilute system and poor for a concentrated one. And the solvent is taken to enter free of solute — for a recycled solvent carrying $x_2$, replace every $y$ with $y - m x_2$ and the equation stays correct.

Finish with the number nobody quotes and everybody needs: these are theoretical stages, and no real column has any. Divide by the overall tray efficiency to get trays to buy. Seventy per cent is a fair absorber assumption; twenty to fifty is common for a poorly soluble gas, which doubles the tower again.