Question

What mass in grams of potassium permanganate is required to make 27.9 mL of a 0.107 M solution?

Answer 1

• Firstly, it's necessary to find an amount of substance of KMnO₄ :

$\nu$(KMnO₄) = $C•V =$ 0.107(M) • 0.0279(ml) = 0.003 mol

• Then the mass of KMnO₄ will be calculated with the formula:

m(KMnO₄) = $\nu • M =$ 0.003(mol) • 158.034(g/mol) = 0.474 g

Answer: m(KMnO₄)=0.474 g