Question #2a118

#"Concentration with respect to calcium hydroxide"= (100xx10^-3*Lxx0.0234*mol*L^-1)/(0.500*L)=??#

Note that #mmol# #=# #10^-3*mol#.

Here's an alternative approach to keep in mind when dealing with dilutions.

As you know, the underlying principle of a dilution is that you can decrease the concentration of a solution by increasing its volume while keeping the number of moles of solute constant.

This implies that increasing the volume of the solution by a factor, let's say #"DF"#, will cause the concentration to decrease by the same factor #"DF"#.

This factor is called dilution factor and can be calculated like this

#color(blue)(|bar(ul(color(white)(a/a)"DF" = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted"color(white)(a/a)|)))#

In your case, you know that the initial volume of the solution, i.e. the volume of the concentrated solution, is equal to #"100 mL"# and that the final volume of the solution, i.e. the volume of the diluted solution, is equal to

#0.500 color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = "500 mL"#

This means that the concentrated solution was diluted by a factor of

#"DF" = (500 color(red)(cancel(color(black)("mL"))))/(100color(red)(cancel(color(black)("mL")))) = color(blue)(5)#

As a result, the concentration of the diluted solution will be #1/color(blue)(5)"th"# the value of the concentration of the concentrated solution

#c_"diluted" = 1/color(blue)(5) * "0.0234 M"#

I tend to think of it like this:

Work out how many moles of solute are contained in the original sample by multiplying the molarity by the number of litres (0.0234 x 0.1) = 0.00234 moles.

The number of moles in the final dilution will be the same, but the total volume will change from 100 ml to 0.5 l .

So the molarity at the end will be 0.00234 / 0.5 which is 0.00468 M.