Living by Chemistry
Living by Chemistry
2nd Edition
ISBN: 9781464142314
Author: Angelica M. Stacy
Publisher: W. H. Freeman
Question
Book Icon
Chapter U4, Problem C15.5RE

(a)

Interpretation Introduction

Interpretation: The steps to make 0.5 M solution of sodium hydroxide needs to be explained.

Concept Introduction: Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

  Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023

(a)

Expert Solution
Check Mark

Answer to Problem C15.5RE

  • Take 19.9 g of solid NaOH solution.
  • Measure 1 L of water with the help of measuring cylinder
  • Add water in beaker containing 19.9 g of solid NaOH
  • Stir the solution till complete NaOH dissolved in it.

Explanation of Solution

To make 0.5 M solution of NaOH; 0.5 M stands for 0.5 moles of NaOH dissolves in 1 L of water.

Molar mass of NaOH = 39.9 g/mol

Calculate the mass of NaOH required:

  Moles = massmolar massMass = moles×molar massMass = 0.5 M × 39.9 g/mol Mass = 19.9 g

Thus to make 0.5 M solution of 1 L of NaOH:

  • Take 19.9 g of solid NaOH.
  • Measure 1 L of water with the help of measuring cylinder
  • Add water in beaker containing 19.9 g of solid NaOH
  • Stir the solution till complete NaOH dissolves in it.

(b)

Interpretation Introduction

Interpretation: The use of 1 L of water while making 0.5 M solution of sodium hydroxide needs to be explained.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

  Moles = massmolar mass1 mole = Na atomsNa = 6.022 × 1023

(b)

Expert Solution
Check Mark

Answer to Problem C15.5RE

The volume of water depends on the volume of solution required. If less than 1 L volume of solution required, less volume of water and corresponding mass of NaOH can be used.

Explanation of Solution

To make 0.5 M solution of NaOH; 0.5 M stands for 0.5 moles of NaOH dissolves in 1 L of water. Thus to make 0.5 M solution of 1 L of NaOH take 19.9 g of solid NaOH and dissolve it in 1 L of water.

Thus the volume of water depends on the volume of solution required. If less than 1 L volume of solution required, less volume of water and mass of NaOH can be used.

Chapter U4 Solutions

Living by Chemistry

Ch. U4.69 - Prob. 3ECh. U4.69 - Prob. 4ECh. U4.69 - Prob. 5ECh. U4.70 - Prob. 1TAICh. U4.70 - Prob. 1ECh. U4.70 - Prob. 2ECh. U4.70 - Prob. 3ECh. U4.70 - Prob. 4ECh. U4.70 - Prob. 6ECh. U4.71 - Prob. 1TAICh. U4.71 - Prob. 1ECh. U4.71 - Prob. 2ECh. U4.71 - Prob. 3ECh. U4.71 - Prob. 4ECh. U4.71 - Prob. 5ECh. U4.71 - Prob. 6ECh. U4.71 - Prob. 7ECh. U4.72 - Prob. 1TAICh. U4.72 - Prob. 1ECh. U4.72 - Prob. 2ECh. U4.72 - Prob. 3ECh. U4.72 - Prob. 4ECh. U4.73 - Prob. 1TAICh. U4.73 - Prob. 1ECh. U4.73 - Prob. 2ECh. U4.73 - Prob. 3ECh. U4.73 - Prob. 5ECh. U4.73 - Prob. 6ECh. U4.73 - Prob. 7ECh. U4.74 - Prob. 1TAICh. U4.74 - Prob. 1ECh. U4.74 - Prob. 2ECh. U4.74 - Prob. 4ECh. U4.74 - Prob. 5ECh. U4.75 - Prob. 1TAICh. U4.75 - Prob. 1ECh. U4.75 - Prob. 2ECh. U4.75 - Prob. 4ECh. U4.75 - Prob. 5ECh. U4.75 - Prob. 6ECh. U4.75 - Prob. 7ECh. U4.75 - Prob. 8ECh. U4.75 - Prob. 9ECh. U4.75 - Prob. 10ECh. U4.75 - Prob. 11ECh. U4.76 - Prob. 1TAICh. U4.76 - Prob. 1ECh. U4.76 - Prob. 2ECh. U4.76 - Prob. 3ECh. U4.76 - Prob. 4ECh. U4.76 - Prob. 5ECh. U4.76 - Prob. 6ECh. U4.76 - Prob. 7ECh. U4.76 - Prob. 8ECh. U4.77 - Prob. 1TAICh. U4.77 - Prob. 1ECh. U4.77 - Prob. 2ECh. U4.77 - Prob. 3ECh. U4.77 - Prob. 4ECh. U4.77 - Prob. 5ECh. U4.77 - Prob. 6ECh. U4.77 - Prob. 7ECh. U4.77 - Prob. 8ECh. U4.78 - Prob. 1TAICh. U4.78 - Prob. 1ECh. U4.78 - Prob. 2ECh. U4.78 - Prob. 3ECh. U4.78 - Prob. 4ECh. U4.78 - Prob. 5ECh. U4.78 - Prob. 6ECh. U4.78 - Prob. 7ECh. U4.78 - Prob. 8ECh. U4.79 - Prob. 1TAICh. U4.79 - Prob. 1ECh. U4.79 - Prob. 2ECh. U4.79 - Prob. 3ECh. U4.79 - Prob. 4ECh. U4.80 - Prob. 1TAICh. U4.80 - Prob. 1ECh. U4.80 - Prob. 2ECh. U4.80 - Prob. 3ECh. U4.80 - Prob. 4ECh. U4.80 - Prob. 5ECh. U4.80 - Prob. 6ECh. U4.80 - Prob. 7ECh. U4.80 - Prob. 8ECh. U4.80 - Prob. 9ECh. U4.80 - Prob. 10ECh. U4.81 - Prob. 1TAICh. U4.81 - Prob. 1ECh. U4.81 - Prob. 2ECh. U4.81 - Prob. 3ECh. U4.81 - Prob. 4ECh. U4.81 - Prob. 5ECh. U4.81 - Prob. 6ECh. U4.81 - Prob. 7ECh. U4.81 - Prob. 8ECh. U4.81 - Prob. 9ECh. U4.82 - Prob. 1TAICh. U4.82 - Prob. 1ECh. U4.82 - Prob. 2ECh. U4.82 - Prob. 3ECh. U4.82 - Prob. 4ECh. U4.82 - Prob. 5ECh. U4.82 - Prob. 6ECh. U4.82 - Prob. 7ECh. U4.82 - Prob. 8ECh. U4.83 - Prob. 1TAICh. U4.83 - Prob. 1ECh. U4.83 - Prob. 2ECh. U4.83 - Prob. 3ECh. U4.83 - Prob. 4ECh. U4.83 - Prob. 5ECh. U4.83 - Prob. 6ECh. U4.83 - Prob. 7ECh. U4.84 - Prob. 1TAICh. U4.84 - Prob. 1ECh. U4.84 - Prob. 2ECh. U4.84 - Prob. 3ECh. U4.84 - Prob. 5ECh. U4.84 - Prob. 6ECh. U4.84 - Prob. 7ECh. U4.85 - Prob. 1TAICh. U4.85 - Prob. 1ECh. U4.85 - Prob. 2ECh. U4.85 - Prob. 3ECh. U4.85 - Prob. 4ECh. U4.85 - Prob. 5ECh. U4.85 - Prob. 6ECh. U4.85 - Prob. 7ECh. U4.85 - Prob. 8ECh. U4.86 - Prob. 1TAICh. U4.86 - Prob. 1ECh. U4.86 - Prob. 2ECh. U4.86 - Prob. 3ECh. U4.86 - Prob. 4ECh. U4.86 - Prob. 6ECh. U4.86 - Prob. 7ECh. U4.86 - Prob. 8ECh. U4.87 - Prob. 1TAICh. U4.87 - Prob. 1ECh. U4.87 - Prob. 2ECh. U4.87 - Prob. 3ECh. U4.87 - Prob. 4ECh. U4.87 - Prob. 5ECh. U4.87 - Prob. 6ECh. U4.87 - Prob. 7ECh. U4.87 - Prob. 8ECh. U4.88 - Prob. 1TAICh. U4.88 - Prob. 1ECh. U4.88 - Prob. 2ECh. U4.88 - Prob. 4ECh. U4.88 - Prob. 5ECh. U4.88 - Prob. 6ECh. U4.88 - Prob. 7ECh. U4.88 - Prob. 8ECh. U4.89 - Prob. 1TAICh. U4.89 - Prob. 1ECh. U4.89 - Prob. 2ECh. U4.89 - Prob. 3ECh. U4.89 - Prob. 4ECh. U4.89 - Prob. 5ECh. U4.89 - Prob. 6ECh. U4.90 - Prob. 1ECh. U4.90 - Prob. 2ECh. U4.90 - Prob. 3ECh. U4.90 - Prob. 4ECh. U4.90 - Prob. 5ECh. U4.90 - Prob. 6ECh. U4.90 - Prob. 7ECh. U4.91 - Prob. 1ECh. U4.91 - Prob. 2ECh. U4.91 - Prob. 3ECh. U4.91 - Prob. 5ECh. U4.91 - Prob. 6ECh. U4.92 - Prob. 1TAICh. U4.92 - Prob. 1ECh. U4.92 - Prob. 2ECh. U4.92 - Prob. 3ECh. U4.92 - Prob. 4ECh. U4.93 - Prob. 1TAICh. U4.93 - Prob. 1ECh. U4.93 - Prob. 2ECh. U4.93 - Prob. 4ECh. U4.93 - Prob. 5ECh. U4.93 - Prob. 6ECh. U4 - Prob. C13.3RECh. U4 - Prob. C13.4RECh. U4 - Prob. C14.1RECh. U4 - Prob. C14.2RECh. U4 - Prob. C14.3RECh. U4 - Prob. C14.5RECh. U4 - Prob. C14.6RECh. U4 - Prob. C15.1RECh. U4 - Prob. C15.2RECh. U4 - Prob. C15.3RECh. U4 - Prob. C15.4RECh. U4 - Prob. C15.5RECh. U4 - Prob. C15.6RECh. U4 - Prob. C15.7RECh. U4 - Prob. C15.8RECh. U4 - Prob. C16.1RECh. U4 - Prob. C16.2RECh. U4 - Prob. C16.3RECh. U4 - Prob. C16.4RECh. U4 - Prob. C17.1RECh. U4 - Prob. C17.2RECh. U4 - Prob. C17.3RECh. U4 - Prob. 1RECh. U4 - Prob. 4RECh. U4 - Prob. 5RECh. U4 - Prob. 6RECh. U4 - Prob. 7RECh. U4 - Prob. 8RECh. U4 - Prob. 9RECh. U4 - Prob. 10RECh. U4 - Prob. 11RECh. U4 - Prob. 12RECh. U4 - Prob. 1STPCh. U4 - Prob. 2STPCh. U4 - Prob. 3STPCh. U4 - Prob. 4STPCh. U4 - Prob. 5STPCh. U4 - Prob. 6STPCh. U4 - Prob. 7STPCh. U4 - Prob. 8STPCh. U4 - Prob. 9STPCh. U4 - Prob. 10STPCh. U4 - Prob. 11STPCh. U4 - Prob. 12STPCh. U4 - Prob. 13STPCh. U4 - Prob. 14STPCh. U4 - Prob. 15STPCh. U4 - Prob. 16STPCh. U4 - Prob. 17STPCh. U4 - Prob. 18STPCh. U4 - Prob. 19STPCh. U4 - Prob. 20STP
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY