Question

The formula weight of an acid is 82 amu. In a titration, $100c{m}^3$ of a solution of this acid containing 39.0 g of the acid per litre was completely neutralized by $95c{m}^3$ of aqueous solution of NaOH containing 40 g of NaOH in 1 l of solution. What is the basicity of the acid?

• A. 4
• B. 2
• C. 1
• D. 3

Answer 1

Answer: (B)

2

Formula weight of acid = 82 amu.
$\therefore$ Molar mass of acid = 82 g

Volume of acid = $100c{m}^3=100ml$
Volume of base $=95c{m}^3=95ml$

$40gNaOH$ per litre

$\therefore N$ of $NaOH=1N$

$39g$ acid per litre

$\therefore$ In 100 ml 3.9 gm acid.

Now milliequivalent of both should be equal-

$\therefore \frac{3.9}{82/x}\times 1000=95\times 1$

$\Rightarrow x=\frac{95\times 82}{3.9\times 1000}=1.997\approx 2$

$\therefore$ n-factor of acid / basicity of acid $=2$.

Hence, option B is correct.