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Chapter 4, Problem 44PS
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### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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Chapter
Section
BuyFindarrow_forward

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# What volume of 2.06 M KMnO4, in liters, contains 322 g of solute?

Interpretation Introduction

Interpretation:

It should be determined that the volume of 2.06MKMnO4 in litres, which contains

322g of solute.

Concept introduction:

• Concentration of solutions can be expressed in various terms; molarity is one such concentration expressing term.
• Molarity (M) of a solution is the number of gram moles of a solute present in one liter of the solution.

Molarity=NumberofmolesofelementVolumeofsolutioninlitres

Molarity=W×1000m×V

Where W, V and m are the mass of solute, volume of solution and molar mass of the solute respectively.

V=W×1000m×Molarity

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
Explanation

The volume of 2.06â€‰Mâ€‰KMnO4 in litres, which contains 322â€‰gâ€‰ of solute is calculated as,

Molarity (M) of a solution is the number of gram moles of a solute present in one litre of the solution.

â€‚Â Molarityâ€‰=Wâ€‰Ã—1000mÃ—V

Where W, V and m are the mass of solute, volume of solution and molar mass of the solute respectively.

The volume of 2.06â€‰Mâ€‰KMnO4 in litres, which contains 322â€‰gâ€‰ of solute can be calculated as follows,

â€‚Â Vâ€‰=

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