Explanation Molarity: The number of moles of solute per liter of solution is known as molarity of the solution. It symbolized by M . Molarity of any solution is a ratio of moles of solutes per liter of solution. It is expressed by the following formula: Molarity ( M ) = moles of solutes liter of solution a. Given: 0.428 L of a 0.450 M K 2 CO 3 solution. Calculation: Step: I To calculate the amount of solutes in the given solution first calculate the number of moles as follows: Molarity ( M ) = moles of solutes liter of solution Moles of solutes = Molarity ( M ) × liter of solution Moles of solutes = 0 .450M × 0 .428 L = 0.1926 moles Step: II Now calculate the amount of solute by their molar mass as follows: Mass of solute = Molar mass × number of moles Mass of solute = 138 .205 g mol − 1 K 2 CO 3 × 0.1926 Moles = 26.6 g K 2 CO 3 b. Given: 10.5 mL of a 2.50 M AgNO 3 solution. Calculation: Step: I First convert mL into L : 1000 mL = 1 .00 L 10.5 mL × 1.00 L 1000 mL = 0.0105 L To calculate the amount of solute in the given solution first calculate the number of moles as follows: Molarity ( M ) = moles of solutes liter of solution Moles of solutes = Molarity ( M ) × liter of solution Moles of solutes = 2

CHEMISTRY: AN INTRODUCTION TO GENERAL, O

13th Edition

ISBN: 9780137444298

Author: Timberlake

Publisher: PEARSON

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Chapter 9, Problem 9.106APP

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To determine: The amount of solute in the given solutions

0.428 L of a 0.450 M K2CO3 solution.
10.5 mL of a 2.50 M AgNO3 solution.
28.4 mL of 6.00 M H3PO4 solution

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Chapter 9, Problem 9.105APP

Chapter 9, Problem 9.107APP

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CHEMISTRY: AN INTRODUCTION TO GENERAL, O

Ch. 9.1 - Prob. 9.1PP Ch. 9.1 - Prob. 9.2PP Ch. 9.1 - Describe the formation of an aqueous KI solution,...Ch. 9.1 - Describe the formation of an aqueous LiBr...Ch. 9.1 - Prob. 9.5PP Ch. 9.1 - Water is a polar solvent and hexane (C6H12)is a...Ch. 9.2 - Prob. 9.7PP Ch. 9.2 - NaOHis a strong electrolyte, and CH3OH is a...Ch. 9.2 - Write a balanced equation for the dissociation of...Ch. 9.2 - Write the balanced equation for the dissociation...

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The amount of solute in the given solutions 0.428 L of a 0.450 M K 2 CO 3 solution. 10.5 mL of a 2.50 M AgNO 3 solution. 28.4 mL of 6.00 M H 3 PO 4 solution

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To determine: The amount of solute in the given solutions 0.428 L of a 0.450 M K2CO3 solution. 10.5 mL of a 2.50 M AgNO3 solution. 28.4 mL of 6.00 M H3PO4 solution

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Chapter 9 Solutions

To determine: The amount of solute in the given solutions

0.428 L of a 0.450 M K2CO3 solution.
10.5 mL of a 2.50 M AgNO3 solution.
28.4 mL of 6.00 M H3PO4 solution