Chapter 8.7, Problem 8.29P

Interpretation Introduction

(a)

Interpretation:

The volume of 6.0 M of NaOH in milliliters is to be determined to prepare the solution of 525 mL of a 2.5 M solution.

Concept Introduction:

The formula that will be used-

  $M_1{V}_1=M_2{V}_2$

Or,

  $V_2=\frac{M_1{V}_1}{M_2}$

Where M represents molarity and V represents volume of the solution.

Interpretation Introduction

(b)

Interpretation:

The volume of 6.0 M of NaOH in milliliters is to be determined to prepare the solution of 750 mL of a 4.0 M solution.

Concept Introduction:

The formula that will be used-

  $M_1{V}_1=M_2{V}_2$

Or,

  $V_2=\frac{M_1{V}_1}{M_2}$

Where M represents molarity and V represents volume of the solution.

Interpretation Introduction

(c)

Interpretation:

The volume of 6.0 M of NaOH in milliliters is to be determined to prepare the solution of 450 mL of a 0.10 M solution.

Concept Introduction:

The formula that will be used-

  $M_1{V}_1=M_2{V}_2$

Or,

  $V_2=\frac{M_1{V}_1}{M_2}$

Where M represents molarity and V represents volume of the solution.

Interpretation Introduction

(d)

Interpretation:

The volume of 6.0 M of NaOH in milliliters is to be determined to prepare the solution of 25 mL of a 3.5 M solution.

Concept Introduction:

The formula that will be used-

  $M_1{V}_1=M_2{V}_2$

Or,

  $V_2=\frac{M_1{V}_1}{M_2}$

Where M represents molarity and V represents volume of the solution.