Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 21, Problem 93A

(a)

Interpretation Introduction

Interpretation: The 250 g of 0.90 % of NaCl solution mass by mass is given. The number of grams of NaCl required to make the solution needs to be determined.

Concept Introduction: In m/m solution, the number of grams of solute in 100 g of the solution is given. Thus, for 0.90 % NaCl m/m solution, there is 0.90 g of NaCl in 100 g of solution.

(a)

Expert Solution
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Explanation of Solution

The mass of the solution is given 250 g; thus, the mass of the solute can be calculated as follows:

For 100 g, 0.90 g of NaCl is used.

For 1 g, the number of grams of NaCl used will be:

  m=0.90100 g

Thus, for 250 g, the mass of NaCl required will be:

  m=2500.90100g=2.25 g

Thus, the mass of NaCl (solute) required will be 2.25 g.

(b)

Interpretation Introduction

Interpretation: The mass of KNO3 required to form the 500 mL of 2.0 M solution needs to be determined.

Concept Introduction: The molarity of a solution is defined as the number of moles of solute in 1 L of the solution.

(b)

Expert Solution
Check Mark

Explanation of Solution

In the problem, 500 mL of 2.0 M KNO3 solution is given. Now, molarity is defined as the number of moles of solute in 1 L of solution.

  M=nVL

Putting the values,

  2.0 M=n500 mL1 L1000 mLOr,n=2.0 M500 mL1 L1000 mL=1 mol

Thus, the number of moles of KNO3 is required. Now, the molar mass of KNO3 is 101.1 g/mol thus, the mass of KNO3 can be calculated as follows:

  m=n×M

Or,

  m=1 mol101.1 g/mol=101.1 g

Therefore, 101.1 g of KNO3 is dissolved in 1 L of water to form the given solution.

Chapter 21 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 21.2 - Prob. 11SPCh. 21.2 - Prob. 12SPCh. 21.2 - Prob. 13SPCh. 21.2 - Prob. 14LCCh. 21.2 - Prob. 15LCCh. 21.2 - Prob. 16LCCh. 21.2 - Prob. 17LCCh. 21.2 - Prob. 18LCCh. 21.2 - Prob. 19LCCh. 21.3 - Prob. 20LCCh. 21.3 - Prob. 21LCCh. 21.3 - Prob. 22LCCh. 21.3 - Prob. 23LCCh. 21.3 - Prob. 24LCCh. 21.3 - Prob. 25LCCh. 21 - Prob. 26ACh. 21 - Prob. 27ACh. 21 - Prob. 28ACh. 21 - Prob. 29ACh. 21 - Prob. 30ACh. 21 - Prob. 31ACh. 21 - Prob. 32ACh. 21 - Prob. 33ACh. 21 - Prob. 34ACh. 21 - Prob. 35ACh. 21 - Prob. 36ACh. 21 - Prob. 37ACh. 21 - Prob. 38ACh. 21 - Prob. 39ACh. 21 - Prob. 40ACh. 21 - Prob. 41ACh. 21 - Prob. 42ACh. 21 - Prob. 43ACh. 21 - Prob. 44ACh. 21 - Prob. 45ACh. 21 - Prob. 46ACh. 21 - Prob. 47ACh. 21 - Prob. 48ACh. 21 - Prob. 49ACh. 21 - Prob. 50ACh. 21 - Prob. 51ACh. 21 - Prob. 52ACh. 21 - Prob. 53ACh. 21 - Prob. 54ACh. 21 - Prob. 55ACh. 21 - Prob. 56ACh. 21 - Prob. 57ACh. 21 - Prob. 58ACh. 21 - Prob. 59ACh. 21 - Prob. 60ACh. 21 - Prob. 61ACh. 21 - Prob. 62ACh. 21 - Prob. 63ACh. 21 - Prob. 64ACh. 21 - Prob. 65ACh. 21 - Prob. 66ACh. 21 - Prob. 67ACh. 21 - Prob. 68ACh. 21 - Prob. 69ACh. 21 - Prob. 70ACh. 21 - Prob. 71ACh. 21 - Prob. 72ACh. 21 - Prob. 73ACh. 21 - Prob. 74ACh. 21 - Prob. 75ACh. 21 - Prob. 76ACh. 21 - Prob. 77ACh. 21 - Prob. 78ACh. 21 - Prob. 79ACh. 21 - Prob. 80ACh. 21 - Prob. 81ACh. 21 - Prob. 82ACh. 21 - Prob. 83ACh. 21 - Prob. 84ACh. 21 - Prob. 85ACh. 21 - Prob. 86ACh. 21 - Prob. 87ACh. 21 - Prob. 88ACh. 21 - Prob. 89ACh. 21 - Prob. 90ACh. 21 - Prob. 91ACh. 21 - Prob. 92ACh. 21 - Prob. 93ACh. 21 - Prob. 94ACh. 21 - Prob. 95ACh. 21 - Prob. 96ACh. 21 - Prob. 97ACh. 21 - Prob. 98ACh. 21 - Prob. 99ACh. 21 - Prob. 100ACh. 21 - Prob. 101ACh. 21 - Prob. 102ACh. 21 - Prob. 103ACh. 21 - Prob. 104ACh. 21 - Prob. 105ACh. 21 - Prob. 106ACh. 21 - Prob. 1STPCh. 21 - Prob. 2STPCh. 21 - Prob. 3STPCh. 21 - Prob. 4STPCh. 21 - Prob. 5STPCh. 21 - Prob. 6STPCh. 21 - Prob. 7STPCh. 21 - Prob. 8STPCh. 21 - Prob. 9STPCh. 21 - Prob. 10STPCh. 21 - Prob. 11STPCh. 21 - Prob. 12STPCh. 21 - Prob. 13STP
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