# Question #7b64f

##### 1 Answer

#### Answer:

#### Explanation:

You are on the right track, but you can solve this one much quicker by using the **dilution factor**.

The **dilution factor** is essentially a measure of how concentrated the initial solution was compared with the *diluted solution*.

Now, the important thing to keep in mind about dilutions is that the amount of solute **must remain constant**. This implies that the **decrease** in concentration must be *matched* by an **increase** in volume.

This is where the dilution factor comes into play

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

In your case, the concentration of the solution must decrease by a factor of

#"DF" = (20color(red)(cancel(color(black)(%))))/(15color(red)(cancel(color(black)(%)))) = color(blue)(4/3)#

This implies that the volume of the solution must **increase** by factor of

#V_"diluted" = "DF" * V_"stock" #

In your case, you will have

#V_"diluted" = color(blue)(4/3) * "1.5 kg" = "2.0 kg"#

Therefore, you can say that adding

#color(darkgreen)(ul(color(black)("mass of water" = "2.0 kg" - "1.5 kg" = "0.50 kg")))#

of water to

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

Let's double-check this result by suing your approach. So, you know that the starting solution contains

Since the diluted solution **must** contain **the same amount** of sulfuric acid, you can say that

In other words, you need to have **for every**

#300 color(red)(cancel(color(black)("g H"_2"SO"_4))) * overbrace("100 g solution"/(15color(red)(cancel(color(black)("g H"_2"SO"_4)))))^(color(blue)("= 15% m/m solution")) = "2000 g solution"#

As you can see, you will once again need to add