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.

Neutralisation of 30 gm of a mixture of acetic acid and phenol solutio

1.	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.
Answer» Solution :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%`