Answer
The molar mass of this acid is equal to 202g/mol.
Work Step by Step
1. Calculate the number of moles of $KOH$:
16.4ml = 0.0164L
$C_1 * V_1 = n(moles)$
$ 0.08133* 0.0164= n(moles)$
$ 0.001334 = n(moles)$
2. Since the acid is monoprotic, to neutralize it, the number of moles of the $KOH$ should be equal.
- n(moles of acid)= $1.334 \times 10^{-3}$
3. Calculate the molar mass:
$MM= \frac{mass(g)}{n(moles)} = \frac{0.2688}{1.334 \times 10^{-3}} =\\ 201.5g/mol$
