Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
Question
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Chapter 13, Problem 54E
Interpretation Introduction

(a)

The molar concentration of 1.00 g KCl in 75.0 mL of solution is to be stated.

Concept introduction:

The molarity is defined as the number of moles present in one liter of solution. General expression is shown below.

Molarity=MolesofsoluteVolumeofsolution

Measuring unit of molarity is mol/L or molar(M).

Expert Solution
Check Mark

Answer to Problem 54E

The molar concentration of 1.00 g KCl in 75.0 mL of solution is 0.178 M.

Explanation of Solution

The molar concentration is calculated by the formula shown below.

M=w(g)×1000Mo(g/mol)×volume of solution (in mL) …(1)

Where,

w is the mass of KCl.

Mois the molar mass of KCl.

M is the molarity of KCl.

The given value of w, Mo and volume of solution is 1.00 g, 74.55 g/mol and 75.0mL respectively.

Substitute all the values of w, Mo and volume of solution in equation 1 as shown below.

M=w(g)×1000Mo(g/mol)×volume of solution (in mL)=1.00 g×100074.55 g/mol×75.0 mL=0.178 M

Therefore, the molar concentration of 1.00 g KCl in 75.0 mL of solution is 0.178 M.

Conclusion

The molar concentration of 1.00 g KCl in 75.0 mL of solution is calculated as 0.178 M.

Interpretation Introduction

(b)

Interpretation:

The molar concentration of 1.00 g Na2CrO4 in 75.0 mL of solution is to be stated.

Concept introduction:

The molarity is defined as the number of moles present in one liter of solution. General expression is shown below.

Molarity=MolesofsoluteVolumeofsolution

Measuring unit of molarity is mol/L or molar(M).

Expert Solution
Check Mark

Answer to Problem 54E

The molar concentration of 1.00 g Na2CrO4 in 75.0 mL of solution is 0.0820 M.

Explanation of Solution

The molar concentration is calculated by the formula shown below.

M=w(g)×1000Mo(g/mol)×volume of solution (in mL) …(1)

Where,

w is the mass of Na2Cr2O7.

Mois the molar mass of Na2CrO4.

M is the molarity of Na2CrO4.

The given value of w, Mo and volume of solution is 1.00 g, 161.97 g/mol and 75.0mL respectively.

Substitute all the values of w, Mo and volume of solution in equation 1 as shown below.

M=w(g)×1000Mo(g/mol)×volume of solution (in mL)=1.00 g×1000161.97 g/mol×75.0 mL=0.0820 M

Therefore, the molar concentration of 1.00 g Na2CrO4 in 75.0 mL of solution is 0.0820 M.

Conclusion

The molar concentration of 1.00 g Na2CrO4 in 75.0 mL of solution is calculated as 0.0820 M.

Interpretation Introduction

(c)

Interpretation:

The molar concentration of 20.0 g MgBr2 in 250.0 mL of solution is to be stated.

Concept introduction:

The molarity is defined as the number of moles present in one liter of solution. General expression is shown below.

Molarity=MolesofsoluteVolumeofsolution

Measuring unit of molarity is mol/L or molar(M).

Expert Solution
Check Mark

Answer to Problem 54E

The molar concentration of 20.0 g MgBr2 in 250.0 mL is 0.434 M.

Explanation of Solution

The molar concentration is calculated by the formula shown below.

M=w(g)×1000Mo(g/mol)×volume of solution (in mL) … (1)

Where,

w is the mass of MgBr2.

Mois the molar mass of MgBr2.

M is the molarity of MgBr2.

The given value of w, Mo and volume of solution is 20.0 g, 184.13 g/mol and 250.0 mL respectively.

Substitute all the values of w, Mo and volume of solution in equation 1 as shown below.

M=w(g)×1000Mo(g/mol)×volume of solution (in mL)=20.0 g×1000184.13 g/mol×250.0 mL=0.434 M

Therefore, the molar concentration of 20.0 g MgBr2 in 250.0 mL is 0.434 M.

Conclusion

The molar concentration of 20.0 g MgBr2 in 250.0 mL is calculated as 0.434 M.

Interpretation Introduction

(d)

Interpretation:

The molar concentration of 20.0 g Li2CO3 in 250.0 mL of solution is to be stated.

Concept introduction:

The molarity is defined as the number of moles present in one liter of solution. General expression is shown below.

Molarity=MolesofsoluteVolumeofsolution

Measuring unit of molarity is mol/L or molar(M).

Expert Solution
Check Mark

Answer to Problem 54E

The molar concentration of 20.0 g Li2CO3 in 250.0 mL of solution is 1.08 M.

Explanation of Solution

The molar concentration is calculated by the formula shown below.

M=w(g)×1000Mo(g/mol)×volume of solution (in mL) … (1)

Where,

w is the mass of Li2CO3.

Mois the molar mass of Li2CO3.

M is the molarity of Li2CO3.

The given value of w, Mo and volume of solution is 20.0 g, 73.89 g/mol and 250.0mL respectively.

Substitute all the values of w, Mo and volume of solution in equation 1 as shown below.

M=w(g)×1000Mo(g/mol)×volume of solution (in mL)=20.0 g×100073.89 g/mol×250.0 mL=1.08 M

Therefore, the molar concentration of 20.0 g Li2CO3 in 250.0 mL of solution is 1.08 M.

Conclusion

The concentration of 20.0 g Li2CO3 in 250.0 mL of solution is calculated as 1.08 M.

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Chapter 13 Solutions

Introductory Chemistry: Concepts and Critical Thinking (8th Edition)

Ch. 13 - Prob. 11CECh. 13 - Prob. 12CECh. 13 - Prob. 1KTCh. 13 - Prob. 2KTCh. 13 - Prob. 3KTCh. 13 - Prob. 4KTCh. 13 - Prob. 5KTCh. 13 - Prob. 6KTCh. 13 - Prob. 7KTCh. 13 - Prob. 8KTCh. 13 - Prob. 9KTCh. 13 - Prob. 10KTCh. 13 - Prob. 11KTCh. 13 - Prob. 12KTCh. 13 - Prob. 13KTCh. 13 - Prob. 14KTCh. 13 - Prob. 15KTCh. 13 - Prob. 16KTCh. 13 - Prob. 17KTCh. 13 - Prob. 18KTCh. 13 - Prob. 19KTCh. 13 - Prob. 20KTCh. 13 - Prob. 1ECh. 13 - Prob. 2ECh. 13 - Prob. 3ECh. 13 - Prob. 4ECh. 13 - Prob. 5ECh. 13 - Prob. 6ECh. 13 - Prob. 7ECh. 13 - Prob. 8ECh. 13 - Prob. 9ECh. 13 - Prob. 10ECh. 13 - Prob. 11ECh. 13 - Prob. 12ECh. 13 - Prob. 13ECh. 13 - Prob. 14ECh. 13 - Prob. 15ECh. 13 - Prob. 16ECh. 13 - Prob. 17ECh. 13 - Prob. 18ECh. 13 - Prob. 19ECh. 13 - Prob. 20ECh. 13 - Prob. 21ECh. 13 - Prob. 22ECh. 13 - Prob. 23ECh. 13 - Prob. 24ECh. 13 - Prob. 25ECh. 13 - Prob. 26ECh. 13 - Prob. 27ECh. 13 - Prob. 28ECh. 13 - Prob. 29ECh. 13 - Prob. 30ECh. 13 - Prob. 31ECh. 13 - Prob. 32ECh. 13 - Prob. 33ECh. 13 - Prob. 34ECh. 13 - Prob. 35ECh. 13 - Prob. 36ECh. 13 - Prob. 37ECh. 13 - Prob. 38ECh. 13 - Prob. 39ECh. 13 - Prob. 40ECh. 13 - Prob. 41ECh. 13 - Prob. 42ECh. 13 - Prob. 43ECh. 13 - Prob. 44ECh. 13 - Prob. 45ECh. 13 - Prob. 46ECh. 13 - Prob. 47ECh. 13 - Prob. 48ECh. 13 - Prob. 49ECh. 13 - Prob. 50ECh. 13 - Prob. 51ECh. 13 - Prob. 52ECh. 13 - Prob. 53ECh. 13 - Prob. 54ECh. 13 - Prob. 55ECh. 13 - Prob. 56ECh. 13 - Prob. 57ECh. 13 - Prob. 58ECh. 13 - Prob. 59ECh. 13 - Prob. 60ECh. 13 - Prob. 61ECh. 13 - Prob. 62ECh. 13 - Prob. 63ECh. 13 - Prob. 64ECh. 13 - Prob. 65ECh. 13 - Prob. 66ECh. 13 - Prob. 67ECh. 13 - Prob. 68ECh. 13 - Prob. 70ECh. 13 - Prob. 71ECh. 13 - Prob. 72ECh. 13 - Prob. 73ECh. 13 - Prob. 74ECh. 13 - Prob. 75ECh. 13 - Prob. 76ECh. 13 - Prob. 77ECh. 13 - Prob. 78ECh. 13 - Prob. 79ECh. 13 - Prob. 80ECh. 13 - Prob. 81ECh. 13 - Prob. 82ECh. 13 - Prob. 83ECh. 13 - Prob. 84ECh. 13 - Prob. 1STCh. 13 - Prob. 2STCh. 13 - Prob. 3STCh. 13 - Prob. 4STCh. 13 - Prob. 5STCh. 13 - Prob. 6STCh. 13 - Prob. 7STCh. 13 - Prob. 8STCh. 13 - Prob. 9STCh. 13 - Prob. 10STCh. 13 - Prob. 11STCh. 13 - Prob. 12STCh. 13 - Prob. 13STCh. 13 - Prob. 14STCh. 13 - Prob. 15STCh. 13 - Prob. 16ST
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