A 0.2688 g sample of a monoprotic acid neutralized 16.4 mL of 0.08133 M KOH solution. Calculate the molar mass of the acid

202 g/mol

Explanation:

Let's consider the neutralization between a generic monoprotic acid and KOH.

HA + KOH → KA + H₂O

The moles of KOH that reacted are:

0.0164 L * 0.08133 mol/L = 1.33 * 10⁻³ mol

The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 * 10⁻³ moles.

1.33 * 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:

0.2688 g/1.33 * 10⁻³ mol = 202 g/mol