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For each compound (all water soluble), would you expect

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

Solution for problem 71E Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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

Chemistry: A Molecular Approach | 3rd Edition

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Problem 71E

For each compound (all water soluble), would you expect the resulting aqueous solution to conduct electrical current?

a. CsCl                 b. CH3OH                 c. Ca(NO2)2                         d. C6H12O6

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to determine the concentration of each ion or molecules in the given set of compounds. Step1: a) NaNO , being an ionic compound will dissociate completely to give sodium and nitrate 3 ion as shown below: NaNO ----3---> Na + NO + 3- + - Here, 0.25 M NaNO will have 0.23 Na ions and 0.25 M NO ions in solution. 3 Step2: b) MgSO , being an ionic compound will dissociate completely to give magnesium and 4 sulfate ion as shown below: 2+ 2- MgSO ----4---> Mg + SO 4 -2 -2 2+ -2 2- Here, 1.3 x 10 M NaNO will have 1.3 x3 M Mg ions and 1.3 x 10 M SO 4 ions in solution. Step3: c) C 6 O12 b6ng a non-ionic compound will not dissociate into ions. Thus, 0.0150 M C H O will6av12.060 M C H O in solution. 6 12 6 Step4: d) Volume of NaCl solution before mixing(V ) = 45.0 mL = 45 / 1000 L = 0.045 L 1 Molarity of NaCl solution before mixing(M ) = 0.272 M 1 Volume of NaCl solution after mixing(V ) = 45.0 + 65.0 = 110.02L = 110/1000 L = 0.11 L Now, using the expression for dilution, we get, V 1M = V1 M 2 2 0.045 X 0.272 = 0.11 X M 2 M 20.01224 / 0.11 M 20.111 M Similarly, For (NH ) CO solution, 4 2 3 Volume of (NH ) CO so4 2n be3re mixing(V ) = 65.0 mL = 65 / 1000 L 10.065 L Molarity of (NH ) CO so4 2n b3ore mixing(M ) = 0.0247 M 1 Volume of (NH ) CO so4 2n af3r mixing(V ) = 45.0 + 65.0 = 110.0 2 = 110/1000 L = 0.11 L Now, using the expression for dilution, we get, V 1M = V 1M 2 2 0.065 X 0.0247 = 0.11 X M 2 M = 0.0016 / 0.11 2 M = 0.0145 M 2 Now, NaCl, being an ionic compound will dissociate completely to give sodium and chloride ions as shown below: + - NaCl ---------> Na + Cl + - Here, 0.111 M NaClwill have 0.111 Na ions and 0.111 M Cl ions in solution. Similarly, (NH ) CO , being an ionic compound will dissociate completely to give 4 2 3 ammonium and carbonate ion as shown below: (NH ) 4 2----3--> 2NH + CO 4 3- + Here, 0.0145 M (NH ) CO will hav4 2x 0.3 5 = 0.029) M NH ions and 0.0145 M 4 2- CO io3 in solution. ------------------

Step 2 of 3

Chapter 4, Problem 71E is Solved
Step 3 of 3

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

Since the solution to 71E from 4 chapter was answered, more than 1989 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. The answer to “?For each compound (all water soluble), would you expect the resulting aqueous solution to conduct electrical current? a. CsCl b. CH3OH c. Ca(NO2)2 d. C6H12O6” is broken down into a number of easy to follow steps, and 25 words. This full solution covers the following key subjects: Solutions, mixture, concentration, concentrations, indicate. This expansive textbook survival guide covers 82 chapters, and 9454 solutions. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. The full step-by-step solution to problem: 71E from chapter: 4 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM.

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