Explanation:
The given data is as follows.
Mass of mixture = 0.3471 g
As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.
This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.
Hence, moles of NaOH in 97 ml = ![\frac{0.1090 \times 97}{1000}](https://tex.z-dn.net/?f=%5Cfrac%7B0.1090%20%5Ctimes%2097%7D%7B1000%7D)
= ![10.573 \times 10^{-3} mol](https://tex.z-dn.net/?f=10.573%20%5Ctimes%2010%5E%7B-3%7D%20mol)
Moles of HCl in 21.00 ml = ![\frac{0.2060 \times 21}{1000}](https://tex.z-dn.net/?f=%5Cfrac%7B0.2060%20%5Ctimes%2021%7D%7B1000%7D)
=
mol
Therefore, total moles of NaOH that reacted are as follows.
-
=
mol
So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.
![\frac{6.247 \times 10^{-3}}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B6.247%20%5Ctimes%2010%5E%7B-3%7D%7D%7B3%7D)
=
mol
Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.
![2.082 \times 10^{-3} mol \times 40 g/mol](https://tex.z-dn.net/?f=2.082%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%5Ctimes%2040%20g%2Fmol)
=
g
= 0.0832 g
Whereas molar mass of benzoic acid is 122 g/mol.
Therefore, 40 g NaOH = 122 g benzoic acid
So, 0.0832 g NaOH = ![\frac{122 g}{40 g} \times 0.0832 g](https://tex.z-dn.net/?f=%5Cfrac%7B122%20g%7D%7B40%20g%7D%20%5Ctimes%200.0832%20g)
= 0.253 g
Hence, calculate the % mass of benzoic acid as follows.
![\frac{0.253 g}{0.3471 g} \times 100](https://tex.z-dn.net/?f=%5Cfrac%7B0.253%20g%7D%7B0.3471%20g%7D%20%5Ctimes%20100)
= 73.10%
Thus, we can conclude that mass % of benzoic acid is 73.10%.