   # Suppose 50.0 mL of 0.250 M CoCl 2 solution is added to 25.0 mL of 0.350 M NiCl 2 solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume that the volumes are additive. ### 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 42E
Textbook Problem
1 views

## Suppose 50.0 mL of 0.250 M CoCl2 solution is added to 25.0 mL of 0.350 M NiCl2 solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume that the volumes are additive.

Interpretation Introduction

Interpretation: The concentrations of the solution for each of ions have to be determined in moles per litre.

Concept Introduction: Concentration of solution can be defined in terms of molarity as moles of solute to the volume of solution. The concentration of solution can be given by,

Concentration(inM)=Molesofsolute(ingrams)Volumeofsolution(inlitres)

### Explanation of Solution

Explanation

• To calculate the moles of CoCl2 and NiCl2

MolCoCl2=0.05000.250molCoCl2L=0.0125molMolNiCl2=0.02500.300molNiCl2L=0.00875mol

0.0125mol Aqueous sample of CoCl2 contains 0.0125mol of Co2+ and 2( 0.0125mol ) = 0.0250mol of Cl- .

Similarly,

0.00875mol Aqueous sample of NiCl2 contains 0.00875mol of Ni2+ and 2( 0.00875mol )= 0.0175molofCl-

Moles of CoCl2 = 0.0125mol

Moles of NiCl2 = 0.00875mol

The moles of CoCl2 and NiCl2 were calculated by plugging in the values of volumes and molarities of their respective solutions. The moles of CoCl2 and NiCl2 were found to be 0

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