   Chapter 15, Problem 84QAP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
1 views

# 84. A solution of the sparingly soluble base Ca(OH)2 is prepared in a volumetric flask by dissolving 5.21 mg of Ca(OH)2 to a total volume of 1000. mL. Calculate the molarity and normality of the solution.

Interpretation Introduction

Interpretation:

The molarity and normality of the given solution is to be calculated.

Concept Introduction:

The atomic mass of an element is defined as the sum of number of protons and number of neutrons. Molar mass of an element is determined from atomic mass of an element.

The number of moles is calculated by the formula,

Moles=MassgMolarmass

The molarity is calculated by the formula,

Molarity=NumberofmolesofsoluteVolumeofsolutionL

The normality of the solution is calculated by the formula,

Normality=Molarity×NumberofH+orOHions.

Explanation

The mass and volume of CaOH2 solution is given to be 5.21mg and 1000.mL.

The conversion of units of 5.21mg into g is done as,

5.21mg=5.211000g=0.00521g

The conversion of units of 1000.mL into L is done as,

1000.mL=1000.1000L=1L

The molar mass of CaOH2 is 74.1g/mol.

The number of moles of CaOH2 in a solution is calculated by the formula,

Moles=MassgMolarmass        (1)

Substitute the values of mass and molar mass in the equation (1).

Moles=0.00521g74.1g/mol=7.03×105moles

The molarity is calculated by the formula,

Molarity=NumberofmolesofsoluteVolumeofsolutionL        (2)

Substitute the values of number of moles of solute and volume of the solution in the equation (2)

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