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
= [tex]0.190 \times 0.024[/tex]
= 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 = [tex]\frac{\text{mass of monoprotic acid}}{\text{no. of moles of acid}}[/tex]
= [tex]\frac{0.410 g}{0.00456 mol}[/tex]
= 89.91 g/mol
Thus, we can conclude that molar mass of the acid is 89.91 g/mol.
