Question

The normality of orthophosphoric acid having purity of 70% by weight and sp.gr. 1.54 would be

(a) 11 N

(b) 22 N

(c) 33 N

(d) 44 N

Answer 1

Correct option is (c) 33 N

Normality = \(\frac{10 \times \text{p(% by mass)} \times \text{d(density g }cm^{-3}) }{\text{Eq. mass}}\)

\(= \frac{10 \times 70 \times 1.54}{98/3} \)

\(= 33\)