10 mL of 0.2 N HCI and 30 mL of 0.1 N HCI together exactly neutralise 40 mL of solution of NaOH, which is also exactly neutralised by a solution in water of 0.61 g of an organic acid. What is the equivalent weight of the organic acid?

It is given that the amount of NaOH neutralizes by 0.61 g of organic acid is equal to the amount of NaOH neutralize by 10 ml of 0.2 N HCl and 30 mL of 0.1 N HCl.

This means that milliequivalents of organic acid is equal to milliequivalent of 10 mL of 0.2 N HCl and milliequivalent of 30 mL of 0.1 N HCl.

$\begin{array}{ccc}\mathrm{Milliequi}. \mathrm{of} \mathrm{HCl}& =& \mathrm{Milliequi}. \mathrm{of} 0.2 \mathrm{N} \mathrm{HCl}+\mathrm{Milliequi}. \mathrm{of} 0.1 \mathrm{N}\hspace{0.17em}\mathrm{HCl}\\ & =& \mathrm{Normality}\times \mathrm{Vol} (\mathrm{in} \mathrm{mL})+\mathrm{Normality}\times \mathrm{Vol} (\mathrm{in} \mathrm{mL})\\ & =& 10 \mathrm{mL}\times 0.2 \mathrm{N}+30 \mathrm{mL}\times 0.1 \mathrm{N}\\ & =& 5\end{array}$Number of equivalents of organic acid is calculated using the mass of organic acid and its equivalent mass because mass of organic acid is provided in the question.

$\mathrm{Equivalents} \mathrm{of} \mathrm{organic} \mathrm{acid}=\frac{\mathrm{Given} \mathrm{mass}}{\mathrm{equivalent} \mathrm{mass}} ....\left(5\right)$

Substitute the values:

$\begin{array}{ccc}5\times {10}^{-3}& =& \frac{\mathrm{Given} \mathrm{mass}}{\mathrm{Equivalent} \mathrm{mass}}\\ \mathrm{Equivalent} \mathrm{mass}& =& \frac{0.61}{5\times {10}^{-3}}\\ & =& 122\end{array}$