Step 1: Set up the rate law for a two reactant reaction.
For $A + B \rightarrow \text{Products}$, the true experimental rate law is usually \[ \text{Rate} = k'[A][B] \] which makes this a second order reaction overall, first order in each reactant.
Step 2: Identify the special condition needed.
If one reactant, say $B$, is present in such a huge excess that its concentration barely changes as the reaction runs, this happens, for instance, when $B$ is the solvent itself, then $[B]$ can be treated as a constant throughout the reaction.
Step 3: Simplify the rate law under that condition.
Folding the near constant $[B]$ into the rate constant gives \[ \text{Rate} = (k'[B])[A] = k_{obs}[A] \] so the reaction now looks and behaves like a first order reaction even though two molecules are genuinely colliding in the mechanism.
Step 4: Give the classic example.
Acid catalyzed hydrolysis of cane sugar, $C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 + C_6H_{12}O_6$, is bimolecular in principle, but since water is present in such vast excess, about $55.5\,M$, the reaction is observed to be pseudo first order in sugar alone. \[ \boxed{\text{One reactant present in large excess}} \]