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

##### 1 Answer

#### Answer:

#### 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**

#color(blue)(ul(color(black)("DF" = c_"stock"/c_"diluted" = V_"diluted"/V_"stock")))#

In your case, you know that the volume of the stock solution is equal to

#"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

#"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**.