Question #a8e62
1 Answer
Explanation:
The thing to remember about dilution calculations is that the ratio that exists between the concentration of the concentrated solution and that of the diluted solution is equal to the ratio that exists between the volume of the diluted solution and of the concentrated solution
#"volume of diluted solution"/"volume of concentrated solution" = "molarity of concentrated solution"/"molarity of diluted solution"#
The factor by which the concentration of the concentrated solution decreased after a dilution is called the dilution factor,
In your case, you know that
#"DF" = (5 color(red)(cancel(color(black)("M"))))/(2.5color(red)(cancel(color(black)("M")))) = color(blue)(2)#
This tells you that the concentration of the concentrated solution was twice as high as the concentration of the diluted solution.
Consequently, the volume of the diluted solution must be twice as large as the volume of the concentrated solution
#"DF" = V_"diluted"/V_"concentrated" implies V_"concentrated" = V_"diluted"/"DF"#
In your case, you have
#V_"concentrated" = "205 mL"/color(blue)(2) = "102.5 mL"#
I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the molarity of the concentrated solution
#color(darkgreen)(ul(color(black)(V_"concentrated" = 1.0 * 10^2"mL")))#
