Question

$\text{The normality of}0.1MB(OH{)}_3$ is

• A. 0.1
• B. 0.9
• C. 0.3
• D. 0.6

Answer 1

The correct option is A 0.1

Boric acid $B(OH{)}_3$ has n-factor 1. Because it does not dissociate to give 3 $O{H}^{-}$ ions, rather it consumes 1 $O{H}^{-}$ ion from the solution to form $B(OH{)}_4^{-}$.

So, $N=M\times n-factor$.
n-factor is 1 for Boric Acid.
Hence, N=0.1N.