# 72.0 ml of a 1.50 M solution is diluted to a volume of 288 mL. A 144 mL portion of that solution...

## Question:

72.0 ml of a 1.50 M solution is diluted to a volume of 288 mL. A 144 mL portion of that solution is diluted using 185 mL of water. What is the final concentration?

## Dilution

The process of dilution involved the addition of water to a concentrated solution to make a dilute solution. This process is governed by the equation

$$M_1V_1 = M_2V_2 $$

where,

- {eq}M_1{/eq} is the intial concentration,

- {eq}V_1{/eq} is the intial volume,

- {eq}M_2{/eq} is the final concentration,

- {eq}V_2{/eq} is the final volume.

## Answer and Explanation: 1

Consider a first cycle of dilution. Use the equation below to solve for the final concentration of the solution. That is,

{eq}M_1V_1 = M_2V_2 {/eq}

where,

- {eq}M_1 = 1.50\ M {/eq} is the intial concentration,

- {eq}V_1 = 72\ mL {/eq} is the intial volume,

- {eq}M_2 {/eq} is the final concentration,

- {eq}V_2 = 288\ mL {/eq} is the final volume.

Putting {eq}M_2 {/eq} on one side of the equation and substituting all given values. Therefore,

{eq}M_2 = \dfrac{M_1V_1}{V_2}\\ M_2 = \dfrac{(1.50\ M)(72.0\ mL)}{288\ mL}\\ M_2 = 0.375\ M {/eq}

Now consider the second cycle of dilution. Solving the final concentration using the same equation.

Where,

- {eq}M_1 = 0.375\ M {/eq} is the intial concentration,

- {eq}V_1 = 144\ mL {/eq} is the intial volume,

- {eq}M_2 {/eq} is the final concentration,

- {eq}V_2 = 144\ mL +185\ mL = 329\ mL {/eq} is the final volume.

Putting {eq}M_2 {/eq} on one side of the equation and substituting all given values. Therefore,

{eq}M_2 = \dfrac{M_1V_1}{V_2}\\ M_2 = \dfrac{(0.375\ M)(144\ mL)}{329\ mL}\\ \boxed{M_2 = 0.1641\ M} {/eq}

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Chapter 4 / Lesson 6Learn about what serial dilution is and its purpose. Understand the serial dilution method, and how to use serial dilution in calculations for microbiology.

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