Can someone please answer this

Question

How many grams of KMnO₄ are required to make 3 L of a 0.25 M solution?

_______ g

Answer 1

Molar mass KMnO4 = 158 g / mole

0.25 M = 0.25 moles / L

0.25 mol / L x 3 L = 0.75 moles required

0.75 moles x 158 g / mol = 118.5 g = 119 g KMnO4 required (3 sig. figs.)