# What weight of potassium permanganate is required to produce 300mL of solution such that 5mL of this solution diluted to 250mL gives a 0.01% w/v solution?

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that you need to start from the diluted solution and work your way back to the original solution.

So, you know that a solution's **mass by volume percent concentration** tells you the number of grams of solute present for every

This means that your diluted solution contains

#250 color(red)(cancel(color(black)("mL solution"))) * "0.01 g KMnO"_4/(100color(red)(cancel(color(black)("mL solution")))) = "0.025 g KMnO"_4#

Now, the trick here is to realize that when you **dilute a solution**, you **decrease** its concentration by keeping the mass of solute **constant** and **increasing** its volume.

This means that when you dilute the **does not change**.

You can thus say that the

So, if you have *the same solution*?

#300 color(red)(cancel(color(black)("mL solution"))) * "0.025 g KMnO"_4/(5color(red)(cancel(color(black)("mL solution")))) = "1.5 g KMnO"_4#

Since you have one **significant figure** for the volume of the initial solution and of the

#color(darkgreen)(ul(color(black)("mass KMnO"_4 = "2 g")))#