Question #03711
1 Answer
Explanation:
This one is a little tricky because it wants you to perform a dilution by adding solute to the solvent, and not the other way around.
Here's how you can approach thus type of problems.
When performing a regular dilution, the dilution factor is used to express the ratio between the concentration of the stock solution and the concentration of the target solution by using the ratio between their two volumes.
Starting from the equation for dilution calculations
#color(blue)(c_1 xx V_1 = c_2 xx V_2)" "# , where
you can rearrange the terms to get
#c_1/c_2 = V_2/V_1#
This is the dilution factor,
#color(blue)("D.F." = V_2/V_1)#
In your case, you want to perform a
#"D.F." = 38 = V_2/V_1#
Now, do not be confused by the fact that you're adding the antibiotic solution to the saline solution!
Let's say that you will end up using a volume of
#V_1 = xcolor(white)(a)"mL"#
Since you're adding this solution to
#V_2 = "7.91 mL" + xcolor(white)(a)"mL" = (7.91 + x)color(white)(a)"mL"#
Plug this into the equation for the dilution factor to get
#38 = V_2/V_1#
#38 = ((7.91 + x)color(red)(cancel(color(black)("mL"))))/(xcolor(red)(cancel(color(black)("mL")))) = (7.91 + x)/x#
This means that you have
#38 * x = 7.91 + x#
#37 * x = 7.91 implies x = 7.91/37 = 0.214#
Therefore, adding
The answer will thus be
#V_"antibiotic" = color(green)("0.214 mL") -># rounded to thre sig figs
So, remember that dilutions are all about keeping the number of moles of solute (or the amount, if you will) constant while increasing the volume of the solution.
So always try to think
final volume
#-># initial volume#-># dilution factor
or
initial concentration
#-># final concentration#-># dilution factor