Question

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?

Answer 1

We need 308.2 grams KMnO4

Step-by-step 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

Answer 2

The answer to your question is 308.1 g of KmNO₄

Step-by-step 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