# You are given a "5-mL" cell suspension with a cell count of 6.5 * 10^5 "cell/mL". You are required to culture the cells at the density of 6.5 * 10^3 "cell/mL". What dilution must you perform to obtain the required density?

Oct 1, 2017

$1 : 100$

#### Explanation:

The problem wants you to figure out how much solvent must be added to your stock solution in order to get the cell count from $6.5 \cdot {10}^{5}$ $\text{cell/mL}$ down to $6.5 \cdot {10}^{3}$ $\text{cell/mL}$.

The idea here is that in any dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution is equal to the dilution factor, $\text{DF}$.

Moreover, the ratio that exists between the volume of the diluted solution and the volume of the stock solution is also equal to the dilution factor.

In your case, the ratio between the two concentrations--stock to diluted-- is

"DF" = (6.5 * 10^5 color(red)(cancel(color(black)("cell/mL"))))/(6.5 * 10^3 color(red)(cancel(color(black)("cell/mL")))) = color(blue)(100)

This means that the ratio between the two volumes--diluted to stock--must also be equal to $\textcolor{b l u e}{100}$.

$\textcolor{b l u e}{100} = {V}_{\text{diluted"/"5 mL}}$

This implies that the volume of the diluted solution will be equal to

${V}_{\text{diluted" = color(blue)(100) * "5 mL" = "500 mL}}$

So in order to get your cell count from $6.5 \cdot {10}^{5}$ $\text{cell/mL}$ to $6.5 \cdot {10}^{3}$ $\text{cell/mL}$, you need to add enough solvent to your stock solution to get the total volume of the diluted solution equal to $\text{500 mL}$.

This will be equivalent to performing a $1 : 100$ dilution, i.e. for every $1$ part of stock solution you get $100$ parts of diluted solution.