<u>Answer:</u> The molar mass of unknown triprotic acid is 97.66 g/mol
<u>Explanation:</u>
To calculate the molarity of acid, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of triprotic acid
are the n-factor, molarity and volume of base which is NaOH.
We are given:
![n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL](https://tex.z-dn.net/?f=n_1%3D3%5C%5CM_1%3D%3FM%5C%5CV_1%3D250mL%5C%5Cn_2%3D1%5C%5CM_2%3D0.0600M%5C%5CV_2%3D95.9mL)
Putting values in above equation, we get:
![3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M](https://tex.z-dn.net/?f=3%5Ctimes%20M_1%5Ctimes%20250%3D1%5Ctimes%200.0600%5Ctimes%2095.9%5C%5C%5C%5CM_1%3D0.0077M)
To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:
![\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BVolume%20of%20solution%20%28in%20mL%29%7D%7D)
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:
![0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol](https://tex.z-dn.net/?f=0.0077M%3D%5Cfrac%7B0.188%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20triprotic%20acid%7D%5Ctimes%20250%7D%5C%5C%5C%5C%5Ctext%7BMolar%20mass%20of%20triprotic%20acid%7D%3D97.66g%2Fmol)
Hence, the molar mass of unknown triprotic acid is 97.66 g/mol