A 0.623 g sample of a monoprotic acid is dissolved in water and titrated with 0.260 M KOH.
What is the molar mass of the acid if 19.0 mL of the KOH solution is required to neutralize the sample?

126. g/mol is the molar mass of the acid .

Explanation:

Moles of KOH = n

Volume of the solution of KOH = 19.0 mL = 0.019 L

Molarity of the KOH solution = 0.260 M

Accrding to recation, 1 mol KOH react with 1 mole of HA ,then 0.00494 mol of KOH will react with :

of HA

Mass of HA = 0.623 g

Molar mass of HA = M

[tex[M=\frac{0.623 g}{0.00494 mol}=126. g/mol[/tex]

126. g/mol is the molar mass of the acid .

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explanation: