Question

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

Answer 1

Answer:

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

$N=3.75\times 10^{-4}N$

Explanation:

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

$Normality=\frac{Number\ of\ equivalent\ weight}{Volume(L)}$

Number of  equivalents weight =

$\frac{Molar\ mass}{n}$

n= acidity of base or basicity of acid

for salts , n = charge present on cation

The relation between Normality and Molarity is :

$Normality =Molarity\times n$

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

$Molarity =\frac{Moles}{Volume}$

$Moles=\frac{mass}{Molar\ mass}$

Molar mass = 400 gram

mass = 50 mg = 0.05 grams

$moles=\frac{0.05}{400}$

$moles=1.25\times 10^{-4}$

$Molarity =\frac{1.25\times 10^{-4}}{Volume}$

let volume = 1 liter

$Molarity =\frac{1.25\times 10^{-4}}{1}$

$molarity=1.25\times 10^{-4}M$

2. Multiply the molarity with n

$Normality =Molarity\times n$

$Normality =1.25\times 10^{-4}M\times n$

n = 3

$Normality =1.25\times 10^{-4}M\times 3$

$Normality =3.75\times 10^{-4}N$