Question

$0.45\text{g}$ of an acid (molar mass: $90\text{g/mol}$) was neutralized by $20\text{mL}$ of a $0.5\text{N}$ caustic potash. The basicity of the acid is:

• A. 1
• B. 3
• C. 2
• D. 4

Answer 1

The correct option is C 2

Let the basicity (n-factor) of the acid be $x$.
$\text{Equivalents of the acid = equivalents of}KOH\text{(caustic potash)}$
$x\times \frac{0.45}{90}=\frac{0.5}{1000}\times 20$
$\Rightarrow x=\frac{0.5}{1000}\times 20\times \frac{90}{0.45}$
On solving, $x=2$
So, the basicity is 2.