Question

4 mol of a mixture of Mohr's salt and Fe2(SO4)3 required 500 ml of 1 M KMnO​4 for complete oxidation in acidic medium. The mole % of Mohr's salt in the mixture is

Answer 1

Normality=Number of gram moles of solute÷Volume of the solution (in liters)
Molarity=Number of gram moles of solute÷Volume of the solution(in liters)
Titration of KMnO4 against Oxalic acid
Preparation of starndards solution of Oxalic acid (500ml M/10(0.1M)solution)
The molecular mass of crystalline oxalic acid is,H2C2O4.2H2O=126
Weight of oxalic acid required to prepare 1000ml of 1M solution=126g
therefore,weight of oxalic acid required to prepare 500ml 0.1M solution=126/1000×500×0.1=6.3g

Answer 2

Answer:

Explanation:

Let no. of moles of Fe(SO4)3 be 'x'.

Then, no. of moles of Mohr's salt is 4-x

Now in the oxidation reaction that takes place, Only Mohr's salt will be oxidized because Fe2(SO4)3 is already oxidized.

This implies that we should mainly consider the Mohr's salt in the reaction that takes place.

In a reaction, No. of equivalents of reactants = No. of equivalents of product.

this implies, No. of equivalents of Mohr's salt = No. of equivalents of KMnO4.

Now, No. of equivalents = No. of moles x n-factor.

No. of Moles of KMnO4 = Vol. of KMnO4 (in litres) X Molarity

Therefore, No. of moles of KMnO4 = 0.2

n-factor of KMnO4 = 5

Therefore No. of equivalents of KMnO4 = 0.2x5 = 1

n-factor of Mohr's salt = |-1| = 1 (n-factor is taken as modulus by formula)

Therefore, No. of equivalents of Mohr's salt = (4-x) x 1

Now, 4 - x = 1

x = 3

Therefore, no. of moles of Fe2(SO4)3 in the mixture = x = 3

And, no. of moles of Mohr's salt in mixture = 1

Now %age of Mohr's salt in mixture = 1/(1+3) x 100 = 1/4 x 100 = 25%.

HOPE IT HELPS!!

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