Chapter 16, Problem 5STP

Interpretation Introduction

Interpretation:

The amount of water added to 6.0 mL of a 0.050M stock solution to dilute it to 0.020M needs to be determined.

Concept introduction:

Process of decreasing concentration of solute in a solution is dilution. It means addition of more solvent without solvent.

Answer to Problem 5STP

B. 9mL is correct option.

Explanation of Solution

Dilution equation is as follows:

${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}$ …… (1)

Here ${\text{M}}_{\text{1}}$ is molarity of stock solution and ${\text{V}}_{\text{1}}$ is volume of stock solution. ${\text{M}}_{\text{2}}$ is molarity of diluted solution and ${\text{V}}_{\text{2}}$ is volume of diluted solution.

Given,

$\begin{array}{c}{\text{M}}_{\text{1}}=0.050\text{M}\\ {\text{V}}_{\text{1}}=6.0\text{mL}\\ {\text{M}}_{\text{2}}=0.020\text{M}\end{array}$

Put this values in equation (1) is as follows:

$\begin{array}{c}{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}\\ \text{0}\text{.05M×6}\text{.0mL}\text{=}\text{0}{\text{.02M×V}}_{\text{2}}\\ {\text{V}}_{\text{2}}\text{=}\frac{\text{0}\text{.05}\bcancel{\text{M}}\text{×6}\text{.0mL}}{\text{0}\text{.02}\bcancel{\text{M}}}\\ \text{=}\text{15mL}\end{array}$

Thus volume of dilute solution is 1.5 mL.

Given that stock solution has volume 6.0 mL. From the calculation diluted solution has volume 15 mL. Thus in order to to dilute it water is added to stock solution is $15\text{mL}-6\text{mL}\text{=}\text{9mL}$. Therefore option B. 9mL is correct.

Conclusion

1. 15 mL

Subtract diluted solution from given stock solution.

2. 6.0 mL

Dilution equation is ${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}$

3. 2.4 mL

Dilution equation is ${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}$