Question

the normality of orthophosphoric acid having purity of 70 % by weight and specific gravity 1.54 is

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:

The normality of the orthophosphoric acid (H₃PO₄) is equal to 33N.

Explanation:

Given that for H₃PO₄ density = (density of water)(specific gravity)

                                                $=1.54*0.998=1.54gml^{-1}$

Mass of H₃PO₄ $=1.54*1000=1540g$

Because only 70% of pure  H₃PO₄

Mass of  H₃PO₄$= 1540*\frac{70}{100}=1078g$  

Equivalent weight of H₃PO₄$=\frac{98}{3} =32.66$ g-eq

The number of gram eq. in H₃PO₄$=\frac{1078}{32.66}=33$

Now Normality = (No. of gram equivalents of H₃PO₄)/Volume( in L)

 ∴ Normality of H₃PO₄ solution  $=\frac{33eq}{1L} =33N$

Therefore  normality for the given solution of  H₃PO₄ with 70% purity  equals to 33N.