52.0 mL of a 1.70 M solution is diluted to a total volume of 258 mL. A 129-mL portion of that...
Question:
52.0 mL of a 1.70 M solution is diluted to a total volume of 258 mL. A 129-mL portion of that solution is diluted by adding 109 mL of water. What is the final concentration? Assume the volumes are additive.
Serial Diluiton
Dilution occurs when solvent is added to a solution with a particular concentration. In adding the solvent, the concentration decreases and the volume increases. When dilution occurs, the new concentration can be determined if the volume is known:
{eq}M_1V_1=M_2V_2 {/eq}
Here the large M stands for concentration in units of molarity and the V is for the volume. State 1 is the concentrated solution and state 2 is the diluted solution.
Answer and Explanation: 1
Here we have 2 diluitons so we will calculate the result of the first dilution and use that to solve for the result of the second dilution.
First we have 52.0mL of a 1.70 M concentrated solution the volume of the diluted solution is 258mL so the new concentration is:
{eq}M_2=\frac{M_1V_1}{V_2}= \frac{1.70M\cdot52.0mL}{258mL}=0.343M {/eq}
Next 129 mL of the 0.343M solution has 109mL water added so that the final volume is 238mL so that the final concentration is:
{eq}M_2=\frac{M_1V_1}{V_2}= \frac{0.343M\cdot129mL}{238mLmL}=0.190M {/eq}
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Calculating Dilution of Solutions
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Chapter 8 / Lesson 5
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