# 75.0 mL of a 1.20 M solution is diluted to a total volume of 248 mL. A 124-mL portion of that...

## Question:

75.0 mL of a 1.20 M solution is diluted to a total volume of 248 mL. A 124-mL portion of that solution is diluted by adding 111 mL of water. What is the final concentration? Assume the volumes are additive.

## Dilution

This process is commonly used by chemists to make a dilute solution. Also, this involved adding water to a concentrated solution to make a dilute solution. This is basically governed by $$M_1V_1 = M_2V_2$$

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.

The final concentration of the solution is determined using the equation below for the first cycle of dilution. That is,

{eq}M_1V_1 = M_2V_2 {/eq}

where,

• {eq}M_1 = 1.2\ M {/eq} is the intial concentration,
• {eq}V_1 = 75.0\ mL {/eq} is the intial volume,
• {eq}M_2 {/eq} is the final concentration,
• {eq}V_2 = 248\ 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{(1.20\ M)(75.0\ mL)}{248\ mL}\\ M_2 = 0.3629\ M {/eq}

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

Where,

• {eq}M_1 = 0.3629\ M {/eq} is the intial concentration,
• {eq}V_1 = 124\ mL {/eq} is the intial volume,
• {eq}M_2 {/eq} is the final concentration,
• {eq}V_2 = 124\ mL +111\ mL = 235\ 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{(0.3629\ M)(124\ mL)}{235\ mL}\\ \boxed{M_2 = 0.1915\ M} {/eq}