# Question #7d6ae

Nov 10, 2016

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

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

It is assumed tha density of vinegar is$= 1 \text{ g/mL}$

So we can say

$100 m L \mathmr{and} 0.1 L$ Vinegar contains $5.1 g$ acetic acid.

So the strength of vinegar solution is $= 5.1 \times 10 = 51 g {L}^{-} 1$

Now molar mass of $C {H}_{3} C O O H$

$= 2 \cdot 12 + 2 \cdot 16 + 4 \cdot 1 = 60 \text{ g/mol}$

So strength of the acetic acid solution

${S}_{1} = \frac{51 \frac{g}{L}}{60 \frac{g}{\text{mol}}} = 0.85 M$

Volume of acetic acid solution

${V}_{1} = 5 m L$

Let this solution is nutralised by $N a O H$ solution of strength ${S}_{2} = 0.22 M$ and of Volume ${V}_{2} m L$

The balanced equation of nutralisation reaction is

$C {H}_{3} C O O H + N a O H \to C {H}_{3} C O O N a + {H}_{2} O$

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

So this follows the relation

${S}_{2} \times {V}_{2} = {S}_{1} \times {V}_{1}$

$\implies 0.22 \times {V}_{2} = 0.85 \times 5$

$\implies {V}_{2} = \frac{0.85 \times 5}{0.22} = 19.3 m L$