Question

The normality of orthophosphoric acid having purity of $70$ % by weight and specific gravity $1.54$ is:

• A. $11N$
• B. $22N$
• C. $33N$
• D. $44N$

Answer 1

Answer: (C)

$33N$

Density of orthophosphoric acid ($H_{3}P{O}_4$) = specific gravity $\times$ density of water

Density of orthophosphoric acid $=1.54\times 0.998=1.54$ g/ml

So mass in $1000$ ml $=1.54\times 1000=1540$ g.
Gram equivalent weight of orthophosphoric acid$=\frac{Molarmass}{N-factor}$ $=\frac{98}{3}=32.66$ g-eq.

As the orthophosphoric acid is only $70$% pure
So the weight of orthophosphoric acid $=1540\times \frac{70}{100}=1078$ g.
Number of gram equivalents of orthophosphoric acid $=\frac{1078}{32.66}=33$

So, the $33$ gram equivalents in $1000$ ml $=33$ N.

as $Normality=\frac{\left(numberofgramequivalents\right)}{\left(volumeofsolutioninL\right)}=33$ N.

Hence, the correct option is $C$