Question

How many grams of KMnO4, are required to prepare 500 ml, 2M solution?

Answer 1

Molar mass of KMnO4 = 158 g.
Molarity = 2 M.
Volume of solution = 500 mL
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Answer 2

Grams of KMnO4, are required to prepare 500 ml, 2M solution is 158 grams.

• KMnO4 is a strong oxidising agent and gets reduced to Mn2+ ion when reacted with a strong reducing agent.
• The Colour of the solution changes from +7 to +2 due to the formation of Mn2+ ions.
• The molecular weight of KMnO4 is 158 grams.
• Molarity =
• $\frac{given \: mass}{molecular \: mass \times volume \: in \: litre}$
• $molarity \times volume \: in \: litre \times \:moleculr \: mass \: = \: mass \: of \: kmno4$
• $So, the \: mass of KMnO4 \: used = 2×158× \frac{500}{1000} = 158 \: gams$
• Henceforth, grams of KMnO4, are required to prepare 500 ml, 2M solution is 158 grams.