Question #ba6e5

1 Answer
Mar 21, 2017


#"7.6% w/v"#


The trick here is to realize that when you dilute a solution, you keep the amount of solute constant and increase the amount of solvent.

Now, you don't actually need to know how much solute you have in your initial solution, all you need to know is the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution.

This ratio will give you the dilution factor, which can be used to determine the ratio that must exist between the concentration of the concentrated solution and the concentration of the diluted solution.

So, you know that you have

#(750 color(red)(cancel(color(black)("mL"))))/(325color(red)(cancel(color(black)("mL")))) = color(blue)(2.3077) -># the dilution factor

So, you know that the volume of the solution increases by a factor of #color(blue)(2.3077)#, which implies that the concentration of the solution must decrease by a factor of #color(blue)(2.3077)#.

This is the case because when we're diluting a solution, we're only increasing the volume of the solvent, meaning that the same amount of solute is now distributed in a larger volume of solution.

You can thus say that

#"% concentrated"/"% diluted" = color(blue)(2.3077)#

which will get you

#"% diluted" = "% concentrated"/color(blue)(2.3077)#

Plug in your value to find

#"% diluted" = "17.5% w/v"/color(blue)(2.3077) = color(darkgreen)(ul(color(black)("7.6% w/v")))#

The answer must be rounded to two sig figs, the number of sig figs you have for the volume of the diluted solution.