Chapter 16.2, Problem 23LC

Interpretation Introduction

Interpretation: The volume of $2.00\text{ M KNO3}$ is to be calculated to prepare $100.0\text{ mL }$ of $0.150\text{ M KNO3}$ .

Concept introduction: Dilution is the process in which the concentrated solution is to be made diluted. This dilution is done by the addition of more solvents in the solution.

Answer to Problem 23LC

The volume of $2.00\text{ M KNO3}$ needed to prepare $100.0\text{ mL }$ of $0.150\text{ M KNO3}$ is $7.5\text{ mL}$ .

Explanation of Solution

In the process of dilution, the number of moles of solute remains constant before and after the dilution. The formula of dilution is derived as below:

  $\text{Molarity = }\frac{n}{V}$

Where:

n is the number of moles of solute.

V is the volume of solution in liters.

As the number of moles of solute remains constant before and after dilution then,

  $n=MV$

Hence, the dilution equation becomes:

  $M\text{1}V\text{1 = }M\text{2}V\text{2}$

In the above formula, the terms are described:

M₁ is the molarity of the stock solution.

V₁ is the volume of stock solution that needs to be taken.

M₂ is the molarity of the new solution.

V₂ is the volume of the new solution.

The data given to find out the volume of stock solution is as follows:

The value of M₁ is $2.00\text{ M}$ .

The value of M₂ is $0.150\text{ M}$ .

The value of V₂ is $100.0\text{ mL}$ .

Substitute the values in the dilution formula.

  $\begin{array}{l}2.00\text{ M }\times V\text{1 = 0}\text{.150 M }\times \text{ 100}\text{.0 mL}\\ V\text{1}=\frac{0.150\text{ mL }\times \text{ 100}\text{.0 mL}}{2.00\text{ M}}\\ V1=7.5\text{ mL}\end{array}$

The volume required is $7.5\text{ mL}$ .

Conclusion

The volume of $2.00\text{ M KNO3}$ needed to prepare $100.0\text{ mL }$ of $0.150\text{ M KNO3}$ is $7.5\text{ mL}$ .