Complete the tableSoluteSolute MassMol SoluteVolume

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo Tro

Problem 69P Chapter 13

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo Tro

Introductory Chemistry | 5th Edition

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Problem 69P

Complete the table

Solute

Solute Mass

Mol Solute

Volume Solution

Molarity

KNO3

22.5 g

_____

125.0 mL

_____

NaHCO3

_____

_____

250.0mL

100 M

C12H22O11

55.38 g

_____

____

0.150 M

Step-by-Step Solution:
Step 1 of 3

Solution 69P:

The Complete table is as follows

Solute

Solute Mass

Mol Solute

Volume Solution

Molarity

KNO3

22.5 g

0.2225 mol

125.0 mL

1.78 M

NaHCO3

2100.0 g

25.0 mol

250.0 mL

100 M

C12H22O11

55.38 g

0.1559 mol

    1.039 L

Or 1039.6 mL

0.150 M

Molar mass of KNO3 = 101.103 g/mol

Mass of KNO3 = 22.5 g

Thus, amount of solute= Mass of solute/molar mass of solute

                                = 22.5 g/101.103 g/mol = 0.2225 mol

Molarity (M) =  =...

Step 2 of 3

Chapter 13, Problem 69P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo Tro
ISBN: 9780321910295

This full solution covers the following key subjects: solute, nahco, mass, mol, molarity. This expansive textbook survival guide covers 19 chapters, and 2045 solutions. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. Since the solution to 69P from 13 chapter was answered, more than 304 students have viewed the full step-by-step answer. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. The full step-by-step solution to problem: 69P from chapter: 13 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. The answer to “Complete the tableSoluteSolute MassMol SoluteVolume SolutionMolarityKNO322.5 g_____125.0 mL_____NaHCO3__________250.0mL100 MC12H22O1155.38 g_________0.150 M” is broken down into a number of easy to follow steps, and 11 words.

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