58.0 mL of a 1.30 M solution is diluted to a total volume 248 mL. A 124-mL portion of that solution is diluted by adding 165 mL of water. What is the final concentration? Assume the volumes are additive
1 Answer
Explanation:
You can solve this problem by using the dilution factor,
The dilution factor is calculated by using either volumes or molarities
#color(blue)(|bar(ul(color(white)(a/a)"DF" = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted"color(white)(a/a)|)))#
So, the dilution factor for the first dilution
#"58.0 mL " -> " 248 mL"#
will be
#"DF"_1 = (248 color(red)(cancel(color(black)("mL"))))/(58.0color(red)(cancel(color(black)("mL")))) = 4.276#
The first solution was diluted by a factor of
#c_"diluted" = c_"concentrated"/"DF"_ 1#
#c_"diluted" = "1.30 M"/4.276 = "0.304 M"#
Now, you take a sample of
#V_"final" = "124 mL" + "165 mL" = "289 mL"#
The concentration of the
The dilution factor for the second dilution will be
#"DF"_ 2 = V_"final"/V_"sample"#
#"DF"_ 2 = (289 color(red)(cancel(color(black)("mL"))))/(124color(red)(cancel(color(black)("mL")))) = 2.331#
Therefore, the concentration of the final solution will be
#c_"final" = c_"sample"/"DF"_ 2#
#c_"final" = "0.304 M"/2.331 = color(green)(|bar(ul(color(white)(a/a)color(black)("0.130 M")color(white)(a/a)|)))#
The answer is rounded to three sig figs.