CHM 111/112 LAB MANUAL >C<
CHM 111/112 LAB MANUAL >C<
11th Edition
ISBN: 9781337310956
Author: SLOWINSKI
Publisher: CENGAGE LEARNING (CUSTOM)
Question
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Chapter 24, Problem 1ASA
Interpretation Introduction

(a)

Interpretation:

The moles of NaOH solution is to be calculated.

Concept introduction:

Number of moles can be calculated as to multiply the molarity by the volume in liters.

It is expressed as,

Moles=Molarity×Volumeofsolutioninliter

Expert Solution
Check Mark

Answer to Problem 1ASA

The moles of NaOH solution is 0.0432 mol.

Explanation of Solution

Volume of initial NaOH solution, (V1) = 7.2 mL= 0.0072 L

Molarity of NaOH solution, (M1) = 6.0 M

Volume of final NaOH solution, (V2) = 400.0 mL= 0.4 L

First we have to find the moles of NaOH, it is calculated as

Moles=Molarity×Volumeofsolutioninliter

Substitute the values of molarity and volume of NaOH in the above expression.

Moles=6.0M×0.0072L=0.0432mol

Conclusion

The number of moles of NaOH is 0.0432 mol.

Interpretation Introduction

(b)

Interpretation:

The molarity of resulting solution is to be calculated.

Concept introduction:

Molarity is defined as number of moles of solute dissolved in 1 liter of solution.

It is expressed as,

Molarity=MolesofsoluteVolumeofsolutioninliter

Expert Solution
Check Mark

Answer to Problem 1ASA

The molarity of resulting solution is 0.108 M.

Explanation of Solution

To find the molarity of NaOH solution, (M2):

Calculate the molarity of NaOH solution using the final volume of NaOH. Molarity=MolesofsoluteVolumeofsolutioninliter

Substitute the values of moles of NaOH and final volume of NaOH solution (V2).

Molarity=0.0432mol0.4L=0.108M

We also calculate the molarity of NaOH solution as

M1V1=M2V2(6.0M)(7.2mL)=M2(400.0mL)M2=0.108M

Therefore, the molarity of resulting solution is 0.108 M.

Conclusion

The molarity of resulting solution is 0.108 M.

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