A chemist needs to prepare 6.00 L of a 0.325 M solution of potassium permanganate. What mass ok KMnO4 Does she need to make the solution?

The answer to your question is 308.1 g of KmNO₄

Explanation:

Data

Volume = 6 L

Concentration = 0.325 M

Molar mass of Potassium permanganate = 39 + 55 + (16 x 4) = 158 g

Process

1.- Calculate the moles of KMnO₄

Formula

Molarity = moles/volume

moles = molarity x volume

moles = 0.325 x 6

moles = 1.95

2.- Calculate the grams of KMnO₄

                      158 g KMnO₄ 1 mol

                         x                   1.95

                         x = (1.95 x 158)/1

                         x = 308.1 g

We need 308.2 grams KMnO4

Explanation:

Step 1: Data given

Volume = 6.00L

Molarity of KMnO4 = 0.325 M

Molar mass KMnO4 = 158.034 g/mol

Step 2: Calculate moles KMnO4

Moles KMnO4 = volume * molarity

Moles KMnO4 = 6.00 L * 0.325 M

Moles KMnO4 = 1.95 moles

Step 3: Calculate mass KMnO4

Mass KMnO4 = moles KMnO4 * molar mass KMnO4

Mass KMnO4 = 1.95 moles * 158.034 g/mol

Mass KMnO4 = 308.2 grams

We need 308.2 grams KMnO4

the answer would be b i believe

i think that the answer is c