EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781118930144
Author: Willard
Publisher: JOHN WILEY+SONS INC.
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Chapter 14, Problem 8PE

(a)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  25.0 g NaNO3+125.0 g H2O

Concept Introduction:

Mass percent is concentration term used to determine concentration of solution in terms of solute percent in particular mass of solution. Expression for mass percent is as follows:

  Mass percent=(Mass of soluteMass of solution)(100)

(a)

Expert Solution
Check Mark

Answer to Problem 8PE

Mass percent of NaNO3 is 16.7 %.

Explanation of Solution

Expression for mass percent of NaNO3 is as follows:

  Mass percent of NaNO3=(Mass of NaNO3Mass of solution)(100)        (1)

Expression to calculate mass of NaNO3 solution is as follows:

  Mass of solution=Mass of NaNO3+Mass of H2O        (2)

Substitute 25.0 g for mass of NaNO3 and 125.0 g for mass of H2O in equation (2).

  Mass of solution=25.0 g+125.0 g=150 g

Substitute 25.0 g for mass of NaNO3 and 150 g for mass of solution in equation (1).

  Mass percent of NaNO3=(25.0 g150 g)(100)=16.7 %

Hence, mass percent of NaNO3 is 16.7 %.

(b)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  1.25 g CaCl2+35.0 g H2O

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 8PE

Mass percent of CaCl2 is 3.45 %.

Explanation of Solution

Expression for mass percent of CaCl2 is as follows:

  Mass percent of CaCl2=(Mass of CaCl2Mass of solution)(100)        (3)

Expression to calculate mass of CaCl2 solution is as follows:

  Mass of solution=Mass of CaCl2+Mass of H2O        (4)

Substitute 1.25 g for mass of CaCl2 and 35.0 g for mass of H2O in equation (4).

  Mass of solution=1.25 g+35.0 g=36.25 g

Substitute 1.25 g for mass of CaCl2 and 36.25 g for mass of solution in equation (3).

  Mass percent of CaCl2=(1.25 g36.25 g)(100)=3.45 %

Hence, mass percent of CaCl2 is 3.45 %.

(c)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  0.75 mol K2CrO4 in 225 g H2O

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 8PE

Mass percent of K2CrO4 is 39.29 %.

Explanation of Solution

Expression to calculate mass of K2CrO4 is as follows:

  Mass of K2CrO4=[(Moles of K2CrO4)(Molar mass of K2CrO4)]        (5)

Substitute 0.75 mol for moles of K2CrO4 and 194.19 g/mol for molar mass of K2CrO4 in equation (5).

  Mass of K2CrO4=(0.75 mol)(194.19 g/mol)=145.64 g

Expression for mass percent of K2CrO4 is as follows:

  Mass percent of K2CrO4=(Mass of K2CrO4Mass of solution)(100)        (6)

Formula to calculate mass of K2CrO4 solution is as follows:

  Mass of solution=Mass of K2CrO4+Mass of H2O        (7)

Substitute 145.64 g for mass of K2CrO4 and 225 g for mass of H2O in equation (7).

  Mass of solution=145.64 g+225 g=370.64 g

Substitute 145.64 g for mass of K2CrO4 and 370.64 g for mass of solution in equation (6).

  Mass percent of K2CrO4=(145.64 g370.64 g)(100)=39.29 %

Hence, mass percent of K2CrO4 is 39.29 %.

(d)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  1.20 mol H2SO4 in 72.5 mol H2O

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Answer to Problem 8PE

Mass percent of H2SO4 is 8.27 %.

Explanation of Solution

Expression to calculate mass of H2SO4 is as follows:

  Mass of H2SO4=[(Moles of H2SO4)(Molar mass of H2SO4)]        (8)

Substitute 1.50 mol for moles of H2SO4 and 39.9 g/mol for molar mass of H2SO4 in equation (8).

  Mass of H2SO4=(1.20 mol)(98.08 g/mol)=117.696 g

Expression to calculate mass of H2O is as follows:

  Mass of H2O=[(Moles of H2O)(Molar mass of H2O)]        (9)

Substitute 72.5 mol for moles of H2O and 18.01 g/mol for molar mass of H2O in equation (9).

  Mass of H2O=(72.5 mol)(18.01 g/mol)=1305.7 g

Expression for mass percent of H2SO4 is as follows:

  Mass percent of H2SO4=(Mass of H2SO4Mass of solution)(100)        (10)

Expression to calculate mass of H2SO4 solution is as follows:

  Mass of solution=Mass of H2SO4+Mass of H2O        (11)

Substitute 117.696 g for mass of H2SO4 and 1305.7 g for mass of H2O in equation (11).

  Mass of solution=117.696 g+1305.7 g=1423.396 g

Substitute 117.696 g for mass of H2SO4 and 1423.396 g for mass of solution in equation (10).

  Mass percent of H2SO4=(117.696 g1423.396 g)(100)=8.27 %

Hence, mass percent of H2SO4 is 8.27 %.

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Chapter 14 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

Ch. 14.5 - Prob. 14.11PCh. 14.5 - Prob. 14.12PCh. 14 - Prob. 1RQCh. 14 - Prob. 2RQCh. 14 - Prob. 3RQCh. 14 - Prob. 4RQCh. 14 - Prob. 5RQCh. 14 - Prob. 6RQCh. 14 - Prob. 7RQCh. 14 - Prob. 8RQCh. 14 - Prob. 9RQCh. 14 - Prob. 10RQCh. 14 - Prob. 11RQCh. 14 - Prob. 12RQCh. 14 - Prob. 13RQCh. 14 - Prob. 14RQCh. 14 - Prob. 15RQCh. 14 - Prob. 16RQCh. 14 - Prob. 17RQCh. 14 - Prob. 18RQCh. 14 - Prob. 19RQCh. 14 - Prob. 20RQCh. 14 - Prob. 21RQCh. 14 - Prob. 22RQCh. 14 - Prob. 23RQCh. 14 - Prob. 24RQCh. 14 - Prob. 25RQCh. 14 - Prob. 26RQCh. 14 - Prob. 27RQCh. 14 - Prob. 28RQCh. 14 - Prob. 29RQCh. 14 - Prob. 30RQCh. 14 - Prob. 31RQCh. 14 - Prob. 32RQCh. 14 - Prob. 33RQCh. 14 - Prob. 34RQCh. 14 - Prob. 35RQCh. 14 - Prob. 37RQCh. 14 - Prob. 38RQCh. 14 - Prob. 39RQCh. 14 - Prob. 40RQCh. 14 - Prob. 41RQCh. 14 - Prob. 42RQCh. 14 - Prob. 1PECh. 14 - Prob. 2PECh. 14 - Prob. 3PECh. 14 - Prob. 4PECh. 14 - Prob. 5PECh. 14 - Prob. 6PECh. 14 - Prob. 7PECh. 14 - Prob. 8PECh. 14 - Prob. 9PECh. 14 - Prob. 10PECh. 14 - Prob. 11PECh. 14 - Prob. 12PECh. 14 - Prob. 13PECh. 14 - Prob. 14PECh. 14 - Prob. 15PECh. 14 - Prob. 16PECh. 14 - Prob. 17PECh. 14 - Prob. 18PECh. 14 - Prob. 19PECh. 14 - Prob. 20PECh. 14 - Prob. 21PECh. 14 - Prob. 22PECh. 14 - Prob. 23PECh. 14 - Prob. 24PECh. 14 - Prob. 25PECh. 14 - Prob. 26PECh. 14 - Prob. 27PECh. 14 - Prob. 28PECh. 14 - Prob. 29PECh. 14 - Prob. 30PECh. 14 - Prob. 31PECh. 14 - Prob. 32PECh. 14 - Prob. 33PECh. 14 - Prob. 34PECh. 14 - Prob. 35PECh. 14 - Prob. 36PECh. 14 - Prob. 37PECh. 14 - Prob. 38PECh. 14 - Prob. 39PECh. 14 - Prob. 40PECh. 14 - Prob. 41PECh. 14 - Prob. 42PECh. 14 - Prob. 44PECh. 14 - Prob. 45PECh. 14 - Prob. 46PECh. 14 - Prob. 47PECh. 14 - Prob. 48PECh. 14 - Prob. 49PECh. 14 - Prob. 50PECh. 14 - Prob. 51PECh. 14 - Prob. 52PECh. 14 - Prob. 53AECh. 14 - Prob. 54AECh. 14 - Prob. 55AECh. 14 - Prob. 56AECh. 14 - Prob. 57AECh. 14 - Prob. 58AECh. 14 - Prob. 59AECh. 14 - Prob. 60AECh. 14 - Prob. 61AECh. 14 - Prob. 62AECh. 14 - Prob. 63AECh. 14 - Prob. 65AECh. 14 - Prob. 66AECh. 14 - Prob. 67AECh. 14 - Prob. 68AECh. 14 - Prob. 69AECh. 14 - Prob. 70AECh. 14 - Prob. 71AECh. 14 - Prob. 72AECh. 14 - Prob. 73AECh. 14 - Prob. 74AECh. 14 - Prob. 75AECh. 14 - Prob. 76AECh. 14 - Prob. 77AECh. 14 - Prob. 78AECh. 14 - Prob. 79AECh. 14 - Prob. 80AECh. 14 - Prob. 81AECh. 14 - Prob. 82AECh. 14 - Prob. 83AECh. 14 - Prob. 84AECh. 14 - Prob. 85AECh. 14 - Prob. 86AECh. 14 - Prob. 87AECh. 14 - Prob. 88AECh. 14 - Prob. 90AECh. 14 - Prob. 91AECh. 14 - Prob. 92AECh. 14 - Prob. 93AECh. 14 - Prob. 94AECh. 14 - Prob. 95AECh. 14 - Prob. 96AECh. 14 - Prob. 97AECh. 14 - Prob. 98AECh. 14 - Prob. 99CECh. 14 - Prob. 100CECh. 14 - Prob. 102CECh. 14 - Prob. 103CECh. 14 - Prob. 104CECh. 14 - Prob. 105CE
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY