   # A stock solution containing Mn 2+ ions was prepaned by dissolving 1.584 g pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions were then prepared by dilution: For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL. For solution B, 10.00 mL of solution A was diluted to 250.0 mL. For solution C, I 0.00 mL of solution B was diluted to 500.0 mL. Calculate the concentrations of the stock solution and solutions A, B, and C. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 6, Problem 44E
Textbook Problem
5 views

## A stock solution containing Mn2+ ions was prepaned by dissolving 1.584 g pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions were then prepared by dilution:For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL.For solution B, 10.00 mL of solution A was diluted to 250.0 mL.For solution C, I 0.00 mL of solution B was diluted to 500.0 mL.Calculate the concentrations of the stock solution and solutions A, B, and C.

Interpretation Introduction

Interpretation: Calculated the concentration of the stock solution and solutions A, B and C.

Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres.  It can be given by the expression,

Weight=volume×molarity×molecularweight1000

### Explanation of Solution

Explanation

Record the given data

Volume and weight of Mn2+ions is =1000mLand1.584g

Molecular weight of Mn2+ions is:54.94 g/mole

The stock solution molarity of Mn2+ ions is calculated as follows,

Weight=volume×molarity×molecularweight1000

1.584g=1000ml×molarity×54.941000

molarity=1.584×10001000×54.94=0.02883M

The stock solution molarity of Mn2+ ions is calculated by plugging the known values in the above equation. The molarity of Mn2+ions isfound to be  0.02883M.

Record the given data

Volume of stock solution of Mn2+ions is =50mL

50mL×1L1000mL×2.833×10-2mole1L=1.442×10-3moleMn2+

Molarity=1.442×10-3mole1000mL×1000mL1L

=1.442×10-3M

The solution A molarity of Mn2+ ions is calculated by plugging the known values in the above equation. The molarity of Mn2+ions isfound to be 1

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
Which of these B vitamins is (are) present only in foods of animal origin? a. niacin b. vitamin c. riboflavin d...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

The senses cannot be completely relied on. Why?

An Introduction to Physical Science

. state the third law of thermodynamics.

Chemistry for Engineering Students

Write a net equation for the photosynthetic process.

Chemistry for Today: General, Organic, and Biochemistry

A wheel 2.00 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular ...

Physics for Scientists and Engineers, Technology Update (No access codes included) 