Explanation The concentration of sodium ion is 0.416 mg/mL. The volume of solution is 0.500 L. From the concentration and volume of solution, the mass of sodium ion in 0.500 L solution can be calculated as follows: C = m V ( L ) On rearranging, m = C × V ( L ) Putting the values, m = ( 0.416 mg/mL ( 10 − 3 L 1 mL ) ) ( 0.500 L ) = 208 mg ( 10 − 3 g 1 mg ) = 0

8th Edition

ISBN: 9780135214756

Author: Robinson

Publisher: PEARSON

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Question

Chapter 4, Problem 4.33P

Interpretation Introduction

Interpretation:

The mass of sodium chloride required to prepare 0.500 L solution needs to be determined.

Concept introduction:

The molarity or concentration of the solution can be calculated as follows:

M=nV(L)

Here, n is number of moles and V is volume of solution in L.

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Chapter 4, Problem 4.32P

Chapter 4, Problem 4.34P

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Chapter 4 Solutions

LCPO CHEMISTRY W/MODIFIED MASTERING

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Knowledge Booster

The mass of sodium chloride required to prepare 0.500 L solution needs to be determined. Concept introduction: The molarity or concentration of the solution can be calculated as follows: M = n V ( L ) Here, n is number of moles and V is volume of solution in L.

Solution Summary: The author explains that the mass of sodium chloride required to prepare 0.500 L solution needs to be determined.

Interpretation:

The mass of sodium chloride required to prepare 0.500 L solution needs to be determined.

Concept introduction:

The molarity or concentration of the solution can be calculated as follows:

M=nV(L)

Here, n is number of moles and V is volume of solution in L.

Want to see the full answer?

Chapter 4 Solutions

The mass of sodium chloride required to prepare 0.500 L solution needs to be determined. Concept introduction: The molarity or concentration of the solution can be calculated as follows: M = n V ( L ) Here, n is number of moles and V is volume of solution in L.

Solution Summary: The author explains that the mass of sodium chloride required to prepare 0.500 L solution needs to be determined.

Interpretation: The mass of sodium chloride required to prepare 0.500 L solution needs to be determined. Concept introduction: The molarity or concentration of the solution can be calculated as follows: M=nV(L) Here, n is number of moles and V is volume of solution in L.

Want to see the full answer?

Chapter 4 Solutions

Interpretation:

The mass of sodium chloride required to prepare 0.500 L solution needs to be determined.

Concept introduction:

The molarity or concentration of the solution can be calculated as follows:

M=nV(L)

Here, n is number of moles and V is volume of solution in L.