Question

45. Calculate the amount of KMnO, required to prepare 100 ml of 0.1 N solution
​

Answer 1

the molecular weight of KMnO4 is about 158 g/mole. Here we want a 0.1 N solution and I am assuming we are using this reagent for its oxidizing power. In a strongly alkaline solution, you get: MnO4¯ + e- → MnO42- So, it only gives up one of it's electrons. Therefore, the normality (N) and molarity (M) are equivalent, providing only one electron. So, if we calculate how much KMnO4 is needed for 100 mL of 0.1 M solution (0.1 mole/L) we will also calculate what is needed for a 0.1 N solution.

100 mL is 0.1 L and 0.1 M is 0.1 moles/liter so we will need

0.1 L * 0.1 mol/L = 0.01 moles.

and, in grams

0.01 moles * 158 g/mole = 1.58 grams

If we were talking about an acid solution, things would be different. In an acidic medium, you get this: MnO4¯ +8H+ +5e- → Mn2+ +4H2O As you can see, Mn gives up 5 electrons. So a 0.1 N solution would correspond to a 0.02 M solution. Running the numbers, ultimately 1/5 as much KMnO4 would be needed or

1.58 g/5 = 0.316 g

Answer 2

Explanation:

this is your answer,...........