   # A solution is prepared by dissolving 0.6706 g oxalic acid (H 2 C 2 O 4 ) in enough water to make 100.0 mL of solution. A 10.00-mL aliquot (portion) of this solution is then diluted to a final volume of 250.0 mL. What is the final molarity of the oxalic acid solution? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 6, Problem 113CWP
Textbook Problem
157 views

## A solution is prepared by dissolving 0.6706 g oxalic acid (H2C2O4) in enough water to make 100.0 mL of solution. A 10.00-mL aliquot (portion) of this solution is then diluted to a final volume of 250.0 mL. What is the final molarity of the oxalic acid solution?

Interpretation Introduction

Interpretation: To calculate the concentration by dissolving 0.6706g oxalic acid in enough water to make 100ml of solution. 10ml aliquot (portion) of this solution is then diluted to a final volume of 250ml . To find final molarity of the oxalic acid solution.

Concept Introduction: By volumetric principle formula is M1×V1=M2×V2

Where M1 = initial concentration of solution

V1 =initial volume of solution

M2 = final concentration of one solution

V2 =Final volume of solution

### Explanation of Solution

Explanation

Given info:

Weight of oxalic acid is 0.6706g , molecular weight of oxalic acid is: 90.03g/mole

0.6706g=100ml×initial concentration×90.03g/mole1000 now rearranged the equation we get

Initial concentration =0.6706g×1000ml100ml×90.03g/mole

Weight of oxalic acid is multiply with  1000ml to give 6.706 ,

100ml is multiply with 90.03g/mole

=6.70690.03

Then 6.706 is divided by 90.03 to give 0.0744M

The final concentration of oxalic acid =0.002976M

According to volumetric principle

M1×V1=M2×V2

Taken 10mlof 0

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