Question

10 ml of 0.2 N $HCl$ and 30 ml of 0.1 N $HCl$ together exactly neutralise 40 ml of a solution of $NaOH$, which is also exactly neutralized by a solution in water of 0.61 g of an organic acid. The equivalent weight of the organic acid is :

• A. 61
• B. 91.5
• C. 122
• D. 183

Answer 1

Answer: (C)

122

$10$ ml of $0.2$ N HCl + $30$ ml of $0.1$ N $HCl=40$ ml of NaOH ($0.61$ g organic acid)

meq. of $HCl=$ meq of $NaOH=meq$ of organic acid

$10\times 0.2+30\times 0.1=\frac{0.61}{E}\times 1000$

$5=\frac{0.61\times 1000}{E}$

$E=\frac{160}{5}=122$

Therefore, the option is C.