Question

Question #7d6ae

Answer 1

The given sample of vinegar is 5.1% acetic acid by mass.

This means `100g` acetic acid contains `5.1g` acetic acid.

It is assumed tha density of vinegar is`=1" g/mL"`

So we can say

`100mL or 0.1L` Vinegar contains `5.1g` acetic acid.

So the strength of vinegar solution is `=5.1xx10=51gL^-1`

Now molar mass of `CH_3COOH`

`=2*12+2*16+4*1=60" g/mol"`

So strength of the acetic acid solution

`S_1=(51g/L)/(60g/"mol")=0.85M`

Volume of acetic acid solution

`V_1=5mL`

Let this solution is nutralised by `NaOH` solution of strength `S_2=0.22M` and of Volume `V_2mL`

The balanced equation of nutralisation reaction is

`CH_3COOH +NaOH->CH_3COONa+ H_2O`

This equation reveals that the stochiometric ratio of reactants is `1:1`

So this follows the relation

`S_2xxV_2=S_1xxV_1`

`=>0.22xxV_2=0.85xx5`

`=>V_2=(0.85xx5)/0.22=19.3mL`