Chapter 4.5, Problem 4.7CYU

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# An experiment calls for you to use 250. mL of 1.00 M NaOH, but you are given a large bottle of 2.00 M NaOH. Describe how to make the desired volume of 1.00 M NaOH.

Interpretation Introduction

Interpretation:

The making of desired volume of 1.00MofNaOH in a large bottle of 2.00MNaOH solution has to be described.

Concept Introduction:

• Concentration of solutions can be expressed in various terms; molarity is one such concentration expressing term.
• Molarity (M) of a solution is the number of gram moles of a solute present in one liter of the solution.

Molarity=MassperlitreMolecular mass

• A solution containing one gram mole or 0.1 gram of solute per litre of solution is called molar solution.
• Concentrationofsubstance=Amountof substancevolumeofthesubstance
• Amountof substance=Concentrationofsubstance×volumeofthesubstance
• V1×M1=V2×M2  where V1andV2 are the initial and final volume of the solution  M1andM2 are the initial and final molar concentration of the solution
Explanation

Measure accurately 125â€‰mL of 2.00â€‰Mâ€‰NaOH into a 250â€‰mL volumetric flask and add water to give a total volume of 250â€‰mL to make 1.00â€‰Mâ€‰ofâ€‰NaOH in a large bottle of 2.00â€‰Mâ€‰NaOH solution.

The initial volume of the solution to make 1.00â€‰Mâ€‰ofâ€‰NaOH in a large bottle of 2.00â€‰Mâ€‰NaOH solution can be determined by using the equation,

â€‚Â V1â€‰Ã—M1â€‰=â€‰V2Ã—M2

Thus,

â€‚Â (2

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