A chemist dissolves 0.9 g of an unknown monoprotic (one acidic H) acid in water. She finds that 14.6 mL of 0.426 M NaOH are required to neutralize the acid.

144.7 g/mol would be the molar mass of the monoprotic acid

Explanation:

We can find the molar mass of the monoprotic acid from the data showed in the excersie. As every titration we can say that:

mEq acid = mEq base

mEq acid = 14.6 mL . 0.426 M

mEq acid = 6.2196 mEq

mEq = Volume (mL) . N = mass (g) / (EQ / 1000), where the EQ is the:

Molar weight / valence. For this case, as the acid is a monoprotic one, we assume Molar Weight = EQ

6.2196 mEq = 0.9 g / (EQ / 1000)

6.2196 mEq . EQ/1000 = 0.9 g

EQ = 0.9 g / 6.2196×10⁻³ = 144.7 g/mol