Answer: The molar mass of monoprotic acid is 135.9 g/mol
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of NaOH solution = 0.1008 M
Volume of solution = 15.0 mL
Putting values in above equation, we get:
[tex]0.1008M=\frac{\text{Moles of NaOH}\times 1000}{15.0}\\\\\text{Moles of NaOH}=\frac{(0.1008\times 15.0)}{1000}=0.00151mol[/tex]
As, the acid is monoprotic, it contains 1 hydrogen ion
1 mole of [tex]OH^-[/tex] ion of NaOH neutralizes 1 mole of [tex]H^+[/tex] ion of monoprotic acid
So, 0.00151 moles of [tex]OH^-[/tex] ion of NaOH will neutralize [tex]\frac{1}{1}\times 0.00151=0.00151mol[/tex] of [tex]H^+[/tex] ion of monoprotic acid
Moles of monoprotic acid = 0.00151 moles
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of monoprotic acid = 0.00151 mole
Given mass of monoprotic acid = 0.2053 g
Putting values in above equation, we get:
[tex]0.00151mol=\frac{0.2053g}{\text{Molar mass of monoprotic acid}}\\\\\text{Molar mass of monoprotic acid}=\frac{0.2053g}{0.00151mol}=135.9g/mol[/tex]
Hence, the molar mass of monoprotic acid is 135.9 g/mol
