# Question a3b28

Feb 4, 2016

$\text{75.22 mL}$

#### Explanation:

I'll stat with a quick rundown of what dilution something actually means.

As you know, the molarity of a solution tells you how many moles of solute you get per liter of solution.

In simple terms, molarity is a measure of how concentrated a solution is in terms of the amount of solute it contains per liter of solution.

When your aim is to dilute a solution, you're essentially keeping the amount of solute constant while increasing the volume of the solution.

This can be written as

$\textcolor{b l u e}{{\overbrace{{c}_{1} \times {V}_{1}}}^{\textcolor{p u r p \le}{\text{moles of solute in initial solution")) = overbrace(c_2 xx V_2)^(color(purple)("moles of solute in target solution}}}}$

Here

${c}_{1}$, ${V}_{1}$ - the concentration and volume of the initial solution
${c}_{2}$, ${V}_{2}$ - the concentration and volume of the target solution

Notice that you can rewrite this equation as

${c}_{1} / {c}_{2} = {V}_{2} / {V}_{1}$

The ratio between the volume of the target solution and the volume of the initial solution gives you the dilution factor.

$\textcolor{b l u e}{\text{D.F". = V_"final"/V_"initial}}$

For example, in order to perform a $1 : 2$ dilution, which implies that the concentration of the target solution is $\frac{1}{2}$ times that of the initial solution, you would use a dilution factor of $2$, since

${c}_{2} = \frac{1}{2} \cdot {c}_{1}$

will give you

${V}_{2} / {V}_{1} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{c}_{1}}}}}{\frac{1}{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{c}_{1}}}}} = 2$

Notice that you don't need concentration values to find the dilution factor of a solution, as is the case with your example.

In your case, you want to perform a $1 : 64$ dilution, which means that the dilution factor will be equal to $64$

$\text{D.F.} = 64 = {V}_{2} / {V}_{1}$

This tells you that the volume of the target solution must be

$64 = {V}_{2} / {V}_{1} \implies {V}_{2} = 64 \cdot {V}_{1}$

Plug in your value for the volume of the aliquot to get

${V}_{2} = 64 \cdot \text{1.194 mL" = "76.416 mL}$

Since the volume of the target solution will be equal to

${V}_{2} = {V}_{1} + {V}_{\text{saline}}$

it follows that you will need to add

V_"saline" = "76.416 mL" - "1.194 mL" = color(green)(color(green)("75.22 mL")#

of sterile saline solution to your anti-cancer drug to perform a $1 : 64$ dilution.