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A 66.7 mL sample of 18.0 M sulfuric acid was diluted with enough water in a volumetric flask to make 5.00 x 102 mL of solution. A 25.0 ml aliquot of this solution was then further diluted to a volume of 1.50 x 102 ml. What is the molarity of the solution after the second dilution?

A 66.7 mL sample of 18.0 M sulfuric acid was diluted with enough water in a volumetric flask to make 5.00 x 102 mL of solution. A 25.0 ml aliquot of this solution was then further diluted to a volume of 1.50 x 102 ml. What is the molarity of the solution after the second dilution?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Current Attempt in Progress A 66.7 mL sample of 18.0 M sulfuric acid was diluted with enough water in a volumetric flask to make 5.00 x 102 mL of solution. A 25.0 mL aliquot of this solution was then further diluted to a volume of 1.50 x 102 mL. What is the molarity of the solution after the second dilution? M eTextbook and Media Hint Save for Later Attempts: 0 of 2 used Submit Answer Type here to search 多 米 insert prt sc %23 %24 & 2 3 4 8 backspace Q W R Y U A D K B N pause M alt alt ctrl
Transcribed Image Text:Current Attempt in Progress A 66.7 mL sample of 18.0 M sulfuric acid was diluted with enough water in a volumetric flask to make 5.00 x 102 mL of solution. A 25.0 mL aliquot of this solution was then further diluted to a volume of 1.50 x 102 mL. What is the molarity of the solution after the second dilution? M eTextbook and Media Hint Save for Later Attempts: 0 of 2 used Submit Answer Type here to search 多 米 insert prt sc %23 %24 & 2 3 4 8 backspace Q W R Y U A D K B N pause M alt alt ctrl
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Step 1

Dilution - In dilution of the any sample the number of moles of the sample will not be changed. Change only in volume.

n1=n2

M1V1=M2V2

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