How much (in #"L"#) of a #"1.2-M"# #"NaCl"# solution must be used to dilute to create #"1 L"# of a #"0.12-M"# #"NaCl"# solution?

Question

Noah Aldrich · Accepted Answer

#"0.1 L"#

The thing to remember about dilution calculations is that the dilution factor, #"DF"#, can be calculated by taking

So for any dilution, you have

#"DF" = color(white)(overbrace(color(black)(c_"stock"/c_"diluted"))^(color(blue)("concentration ratio: stock/diluted"))) = color(white)(overbrace(color(black)(V_"diluted"/V_"stock"))^(color(blue)("volume ratio: diluted/stock")))#

In your case, the dilution factor is equal to

#"DF" = (1.2 color(red)(cancel(color(black)("M"))))/(0.12color(red)(cancel(color(black)("M")))) = color(blue)(10)#

This tells you that the volume of the diluted solution is #color(blue)(10)# times the volume of the stock solution.

You can thus say that you have

#V_"stock" = V_"diluted"/"DF"#

which, in your case, will get you

#V_"stock" = "1 L"/color(blue)(10) = color(darkgreen)(ul(color(black)("0.1 L")))#

The answer is rounded to one significant figure.

So, in order to perform this dilution, you take #"0.1 L"# of a #"1.2-M"# stock solution of sodium chloride and add enough water to get the final volume of the solution to #"1 L"#.

This will decrease the concentration of sodium chloride from #"1.2 M"# in the stock solution to #"0.12 M"# in the diluted solution.

Axel Alvarado · Answer

undefined To dilute the solution, you need to use $ \frac{0.12}{1.2} $ liters of the $1.2 \text{ M}$ solution, which is $0.1 \text{ L}$.