Question

Which among the following has the highest normality?

• A. 8 g of KOH per 100 ml of solution
• B. 0.5 M sulphuric acid
• C. 1 N phosphoric acid
  
• D. 2 moles of NaOH per 1000 ml of solution

Answer 1

The correct option is D 2 moles of NaOH per 1000 ml of solution

$N=\frac{\text{no.of equivalents}}{\text{Volume of solution(L)}}$
N of KOH = $\frac{\left[\frac{8}{56}\right]}{0.1}=1.43N$
N of $H_{2}S{O}_4$ = M $\times$ n factor = $0.5\times 2=1N$
equivalents of NaOH = moles $\times$ n factor
= $2\times 1=2$eq
N of NaOH = $\frac{2}{1}=2$N