# 300 mL of water is added to 5 mL of a 12 M solution of "KI". What is the resulting molarity of the solution?

Feb 21, 2017

$\text{0.2 M}$

#### Explanation:

The trick here is to realize that in a dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution is equal to the ratio that exists between the volume of the diluted solution and the volume of the stock solution.

These two ratios give you the dilution factor

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{DF" = c_"stock"/c_"diluted" = V_"diluted"/V_"stock}}}}$

In your case, you know that the volume of the stock solution is equal to $\text{5 mL}$. Adding $\text{300 mL}$ of water to the stock solution will dilute it to a final volume of

$\text{5 mL " + " 300 mL" = "305 mL}$

This means that the dilution factor will be equal to

"DF" = (305 color(red)(cancel(color(black)("mL"))))/(5color(red)(cancel(color(black)("mL")))) = color(blue)(61)

Therefore, you can say that the concentration of the diluted solution will be

$\text{DF" = c_"stock"/c_"diluted" implies c_"diluted" = c_"stock"/"DF}$

Plug in your values to find

c_"diluted" = "12 M"/color(blue)(61) = color(darkgreen)(ul(color(black)("0.2 M")))

The answer must be rounded to one significant figure.