   Chapter 15, Problem 138CP ### 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

# A solution is prepared by dissolving 0.6706 g of 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:

The final molarity of the oxalic acid 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.

Explanation

The mass of oxalic acid and volume of solution is given to be 0.6706g and 100.0mL respectively.

The molar mass of oxalic acid is 90.03g/mol.

The number of moles of initial solution is calculated by the formula,

Moles=MassgMolarmass    (1)

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

Moles=0.6706g90.03g/mol=0.00745mol

The solution is diluted to a final volume of 250.0mL.

The conversion of units of 250.0mL into L is done as,

250.0mL=250.01000L=0.25L

It is given that only 10

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