Question

When 3.92g/l of a sample of mohr's salt react completely with 50ml.N/10 kmno4 solution. The percentage purity of the sample of mohr's salt is?

Answer 1

Hey mate here's your answer

Step 1:

Equivalent weight of ferrous ammonium sulphate =392

∴ normality of salt solution =3.92392×1000100

⇒0.1N

V1N1=V2N2

V1N1⇒Mohr's salt

V2N2⇒KmnO4

20×0.1=18×N2

N2=20×0.118

⇒19N

Step 2:

Equivalent weight of KMno4=31.6

∴ weight of KMnO4=31.6

∴ Weight of KMnO4 in litre=31.6×19

⇒3.5g

Answer 2

Answer:

50%

Explanation:

Molecular weight of Mohr's salt (Ferrous Ammonium Sulphate) is 392 g/mole

Considering the solution to be 1 Litre and its weight to be x gram

Normality of the Mohr's Salt N₁ = (x/392) × (1/1) = x/392

It reacts with 50 mL N/10 KMnO₄

Thus,

using N₁V₁ = N₂V₂

$\frac{x}{392}\times 1000=50\times \frac{1}{10}$

$\implies x=0.392\times5=1.96$

Thus in 3.92 g, the actual Mohr's salt is only 1.96 g

%purity = $\frac{1.96}{3.92}\times 100$

            = 50 %

Hope this is helpful.