Answer:
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
