Answer:
[tex]\boxed{\text{(a) 0.001 66 mol MOH; (b) 200 g/mol}}[/tex]
Explanation:
I am assuming that the unknown acid is monobasic. Then the general equation for the reaction is
HA + MOH ⟶ MA + H₂O
(a) Moles of MOH
[tex]\text{Moles of MOH}= \text{0.008 72 L MOH} \times \dfrac{\text{0.19 mol MOH}}{\text{1 L MOH}} = \text{0.001 66 mol MOH}[/tex]
(b) Moles of HA
Now, you use the molar ratio from the balanced chemical equation to find the moles of unknown acid.
[tex]\text{Moles of HA = 0.001 66 mol MOH} \times \dfrac {\text{1 mol HA}}{\text{1 mol MOH }} = \text{0.001 66 mol HA}[/tex]
(c) Molar mass of HA
[tex]\text{MM} =\dfrac{\text{mass}}{\text{moles}}= \dfrac{\text{0.33 g}}{\text{0.001 66 mol}} = \text{200 g/mol}[/tex]
The molar mass of the unknown acid is [tex]\boxed{\textbf{200 g/mol}}[/tex]
