Neutralisation of 30 gm of a mixture of acetic acid and phenol solutions required 100 ml of 2M sodium hydroxide solution. When the same mixture was treated with bromine water, `33.1` gm of precipitate was formed. Determine the mass percentage of acetic acid and phenol in the given solution.

Explanation

i.
331 gm of (B) is obtained form 94 gm of (A).
`33.1` gm (B) is obtained form `= (94)/(331)xx33.1`
`= 9.4` gm of phenol
Weight of phenol `=9.4` gm
`= (9.4)/(9.4)= 0.1` mol
ii. NaOH will react with both `CH_(3)COOH` and phenol. Total molar equivalent of NaOH `= 100xx2`
`= 200` mEq
`= (200)/(1000)= 0.2` Eq. .of NaOH
`= 0.2` mol of NaOH
Acid + Phenol `= 0.2` mol
Acid + `0.1` mol `= 0.2` mol
`:.` Acid `= 0.2 - 0.1 = 0.1` mol
`=0.1` Eq.
1 Eq. of `CH_(3)COOH = 60` gm
`0.1` Eq. of `CH_(3)COOH = 6` gm
Weight of acid = 6 gm
Weight of phenol `= 9.4` gm
Mass percentage of acid `= (6)/(30)xx100 = 20%`
Mass percentage of phenol `= (9.4)/(30)xx100=31.3%`