Question

A water sample contains 50Mg of ferric sulfate. determine the normality of the solution

Answer 1

Assuming Volume of dolution = 1 liter . The Normality is:

`N=3.75* 10^(-4)N`

Step-by-step explanation:

Normality : It is the gram equivalent of solute per liter of solution. It is represented by N.

`Normality=(Number\ of\ equivalent\ weight)/(Volume(L))`

Number of equivalents weight =

`(Molar\ mass)/(n)`

n= acidity of base or basicity of acid

for salts , n = charge present on cation

The relation between Normality and Molarity is :

`Normality =Molarity* n`

n = charge present on cation (not moles)

`Fe_(2)(SO_(4))_(3)` = ferric sulphate

Here Fe is cation and its oxidation state is = + 3 so n= 3

1. First , calculate the molarity(M)

Molar Mass of ferric sulfate = Fe2(SO4)3

2(mass of Fe)+3 (mass of S) + 12(mass of O)

= 2(56)+3(32)+12(16)

= 400 grams

`Molarity =(Moles)/(Volume)`

`Moles=(mass)/(Molar\ mass)`

Molar mass = 400 gram

mass = 50 mg = 0.05 grams

`moles=(0.05)/(400)`

`moles=1.25* 10^(-4)`

`Molarity =(1.25* 10^(-4))/(Volume)`

let volume = 1 liter

`Molarity =(1.25* 10^(-4))/(1)`

`molarity=1.25* 10^(-4)M`

2. Multiply the molarity with n

`Normality =Molarity* n`

`Normality =1.25* 10^(-4)M* n`

n = 3

`Normality =1.25* 10^(-4)M* 3`

`Normality =3.75* 10^(-4)N`