×
Log in to StudySoup
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 76e
Join StudySoup for FREE
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 76e

Already have an account? Login here
×
Reset your password

Solved: Complete and balance each equation. If no reaction

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 76E Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

4 5 1 365 Reviews
31
0
Problem 76E

Problem 76E

Complete and balance each equation. If no reaction occurs, write “NO REACTION.”

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the mass of solid silver nitrate required to prepare 200.0 mL of 0.150 M silver nitrate solution. Also, we are going to calculate the volume of 3.6 M nitric acid required to prepare 100 mL of 0.50 M nitric acid solution. Step1: Molarity is defined as the number of moles of the solute(Here silver nitrate is the solute) in 1 litre of the solution. Mathematically, Number of moles of solute Molarity(M) = -------------------------------------- ------(1) Volume of solution in litres Step2: a) Given, Desired volume of the silver nitrate solution = 200 mL = 200 / 1000 L = 0.200 L Molarity of the silver nitrate solution = 0.150 M Now, using the molarity expression, we get, Number of moles of AgNO = volume o3solution in litres X Molarity AgNO3 = 0.200 X 0.150 M = 0.03 mol Now, 1 mole of AgNO = molar mass of AgNO 3 3 Therefore, 0.03 moles of AgNO = 0.03 X 3lar mass of AgNO 3 = 0.03 X 169.87 g [molar mass of AgNO = 169.87 g/mol] 3 = 5.0961 g Thus, the required weight of silver nitrate is 5.0961 g. Step3: In order to dilute a stock solution, we have to use the following dilution equation, M 1 1M V2 2 -----------(2) Here, M and V are the molarity and volume of the concentrated stock solution whereas M and 1 1 2 V2re the molarity and volume of the diluted solution. b) Again, for the second problem, let us consider the 3.6 M solution of nitric acid to be concentrated solution and the solution of nitric acid to be prepared as dilute solution. Therefore, M (co1entrated) = 3.6 M V (concentrated) = 1 M 2ilute) = 0.50 V (dilute) = 100 mL 2 Putting these values in the equation(2), we get, 3.6 M X V = 010 M X 100 mL 3.6 M X V = 51(M x mL) V1 50 / 3.6 [(M x mL)/ M] V1 13.89 mL Thus, the required volume of nitric acid is 13.89 mL Again, The volume of water required = Final volume of the solution - volume of nitric acid used = 100 mL - 13.89 mL = 86.11 mL Thus, the required volume of water is 86.11 mL. ----------------

Step 2 of 3

Chapter 4, Problem 76E is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Solved: Complete and balance each equation. If no reaction