# Question #a8e62

##### 1 Answer

#### Answer:

#### Explanation:

The thing to remember about **dilution calculations** is that the **ratio** that exists between the **concentration** of the concentrated solution and that of the diluted solution is **equal** to the ratio that exists between the **volume** of the diluted solution and of the concentrated solution

#"volume of diluted solution"/"volume of concentrated solution" = "molarity of concentrated solution"/"molarity of diluted solution"#

The factor by which the concentration of the concentrated solution **decreased** after a dilution is called the **dilution factor**,

In your case, you know that

#"DF" = (5 color(red)(cancel(color(black)("M"))))/(2.5color(red)(cancel(color(black)("M")))) = color(blue)(2)#

This tells you that the concentration of the concentrated solution was **twice as high** as the concentration of the diluted solution.

Consequently, the **volume** of the diluted solution must be **twice as large** as the volume of the concentrated solution

#"DF" = V_"diluted"/V_"concentrated" implies V_"concentrated" = V_"diluted"/"DF"#

In your case, you have

#V_"concentrated" = "205 mL"/color(blue)(2) = "102.5 mL"#

I'll leave the answer rounded to two **sig figs**, but keep in mind that you only have one significant figure for the molarity of the concentrated solution

#color(darkgreen)(ul(color(black)(V_"concentrated" = 1.0 * 10^2"mL")))#