Respuesta :
Answer: The molar mass of unknown triprotic acid is 97.66 g/mol
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of triprotic acid
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL[/tex]
Putting values in above equation, we get:
[tex]3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M[/tex]
To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of solution = 0.0077 M
Given mass of triprotic acid = 0.188 g
Volume of solution = 250 mL
Putting values in above equation, we get:
[tex]0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol[/tex]
Hence, the molar mass of unknown triprotic acid is 97.66 g/mol
