Question

suppose we are preparing a solution of potassium permanganate KMno⁴ and need about 0.10 mol of the compound (that is 0.10 mol KMno⁴) How many grams of the compound do we need

Answer 1

Let the weight required be x g.Molarity c=0.1 MNumber of moles of KMnO4​ is  n=158.034
x​ moles
The volume of solution in litrs is V=1000
250​=0.25 litresMolarity of the given solution is given as c=V
n​c=0.25
158.034
x​​=0.1⟹x=158.034×0.1×0.25⟹x=3.95 gTherefore, weight of KMnO4​ required is 3.95 g.