Question

Calculate Normality of a solution containing 15.8g of KMnO4 in 50mL acidic solution. ​

Answer 1

normality=15.8/158×1000/50=2normality is your answer

Answer 2

$\huge\underline\purple{\sf Answer:-}$

★$\large{\boxed{\sf Normality (N)=10}}$

$\huge\underline\purple{\sf Solution:-}$

We know that :-

★$\large{\boxed{\sf Normality (N)={\frac{W×1000}{E×V_{mL}}}}}$

★Given :-

Mass (W) = 15.8g

Volume (V)= 50mL

Equivalent (E)=?

Normality (N) = ?

★$\large{\boxed{\sf E={\frac{Molar\:Mass\:Of\:KMnO_{4}}{Valence\:Factor}}}}$

On putting value :-

$\large\implies{\sf {\frac{158}{5}}}$

$\large{\boxed{\sf E =31.6}}$

So ,

$\large{\sf N ={\frac{15.8×1000}{31.6×50}}}$

$\large\implies{\sf {\frac{15800}{1580}}}$

★$\large\red{\boxed{\sf Normality ( N) =10}}$