Answer:
12.1-M requires less volume of stock solution.
Explanation:
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In this case, according to the foundations of dilution processes, it turns out convenient to remember that the moles of the solute before and after the dilution remains the same; for that reason we use the following equation:
[tex]M_1V_1=M_2V_2[/tex]
Relating the initial and final molarities and volumes. Now, it is seen that V2 is 10.0 mL and M2 is 1.0 M, which means that if we solve for V1 for 12.1 M and 6.0 M, we obtain the following volumes:
[tex]V_1=\frac{M_2V_2}{M_1} \\\\V_1^{12.1M}=\frac{1.0M*10.0 mL}{12.1M}=0.826mL\\\\V_1^{6.0M}=\frac{1.0M*10.0 mL}{6.0M}=1.67mL[/tex]
Therefore, since the 12.1-M solution requires less volume of stock solution, it makes sense to use this one as the initial solution.
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