Question

How many grams of KMnO4 are required to prepare 1.0 L of a solution of 1.5 M KMNO4?

Answer 1

Molarity=No of moles of solute(KMnO₄)/Volume of solution(L)

Given values:
Molarity=1.5
Volume of solution=1 L

Lets find No of moles of KMnO₄:

1.5= No of moles of KMnO₄/1

No of moles=1.5

Moles=given mass/Molar mass

Molar mass of KMNO₄=158

Here we need to find the given mass

1.5=given mass/158
given mass=1.5×158
                  =237g

Hence we require 237 g of KMnO₄ to make a solution of 1.5 M