# 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**, **decreased** upon dilution.

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 **decreased** 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 **add** another **final volume** of the solution will be

#V_"final" = "124 mL" + "165 mL" = "289 mL"#

The concentration of the **equal** to the concentration of the first diluted solution, i.e.

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**.