Question

Normality of 10% acetic acid is _______.

Answer 1

Step 1: Finding the molecular mass of Acetic acid

Basically, 10 percent of the $\frac{\mathrm{w}}{\mathrm{v}}$ Acetic acid will imply that $10\mathrm{grams}$ of the Acetic acid is dissolved in the $100\mathrm{ml}$ of water.

The molar mass of ${\mathrm{CH}}_3\mathrm{COOH}$ = $$\begin{gathered} 2\times 12+4\times 1+2\times 16 \\ =60\mathrm{g} \end{gathered}$$.

So, one mole of ${\mathrm{CH}}_3\mathrm{COOH}$ = $60\mathrm{g}\mathrm{by}\mathrm{mass}$.

Therefore, $100\mathrm{g}\mathrm{of}{\mathrm{CH}}_3\mathrm{COOH}\mathrm{will}\mathrm{give}=\frac{100}{60}=1.67\mathrm{moles}$.

Step 2: Finding the Normality of the $10\%$ Acetic acid

$\mathrm{Normality}=\mathrm{Number}\mathrm{of}\mathrm{gram}\mathrm{equivalents}\times {[\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{litres}]}^{-1}$.

$1000\mathrm{ml}\mathrm{of}\mathrm{the}\mathrm{solution}\mathrm{having}100\mathrm{g}\mathrm{of}\mathrm{the}\mathrm{solute}$

Therefore, $\frac{1000}{60}=1.67\mathrm{N}$

Hence, the normality of 10% acetic acid is $1.67\mathrm{N}$.