Question

The normality of orthophosphoric acid having purity of 70% will be

Answer 1

Mass =70 %*1.54 =1.078 g/ml

Mass in 1000 ml solution=1.078*1000=1078 g

Molar mass of orthophosphoric acid=98 g

Normality of orthophosphoric acid=(Mass of solute in 1L)/(Molar mass/basicity)

=(1.078)/(98/3)

=33 N

So,the normality of orthophosphoric acid having purity of 70 % by weight and specific gravity 1.54 is 33N.

Answer 2

Answer:

• $33N$

Explanation:

$N =\dfrac{x\% \times sp.gr\times 10}{GEW}$

$N = \dfrac{70\times 1.54\times 10\times 3}{98}$

$N = \dfrac{5\times 1.54\times 10\times 3}{7}$

$N = \dfrac{ 7.7 \times 10 \times 3}{7}$

$N = \dfrac{77 \times 3}{7}$

$N = \cancel{\dfrac{231}{7}}$

$N = 33$