Question

A 0.410 g sample of a monoprotic acid is dissolved in water and titrated with 0.190 M KOH . What is the molar mass of the acid if 24.0 mL of the KOH solution is required to neutralize the sample?

Answer 1

Answer: Molar mass of the acid is 89.91 g/mol.

Explanation:

The given data is as follows.

    Weight of monoprotic acid = 0.410 g

    Molarity of KOH = 0.190 M

    Volume of KOH used = 24 ml = 0.024 L    (as 1 ml = 0.001 L)

First, we will calculate the number of moles of KOH as follows.

    No. of moles = Molarity × Volume

                        = $0.190 \times 0.024$  

                        = 0.00456 mol

So, using KOH the given monoprotic acid is neutralized in 1:1 ratio. Therefore,

  moles of monoprotic acid = moles of KOH

                                          = 0.00456 mol

Hence, we will calculate the molar mass as follows.

     Molar mass = $\frac{\text{mass of monoprotic acid}}{\text{no. of moles of acid}}$

                         = $\frac{0.410 g}{0.00456 mol}$

                         = 89.91 g/mol

Thus, we can conclude that molar mass of the acid is 89.91 g/mol.