Question

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?

Answer 1

`"2 g"`

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 `"100 mL"` of the solution.

This means that your diluted solution contains `"0.01 g"` of potassium permanganate for every `"100 mL"` of the solution. Consequently, you can say that the `"250-mL"` sample of this 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 `"5-mL"` sample to `"250 mL"`, the mass of potassium permanganate does not change.

You can thus say that the `"5-mL"` sample contained `"0.025 g"` of potassium permanganate, the amount of solute present in the diluted solution.

So, if you have `"0.025 g"` of solute in `"5 mL"` of the solution, how many grams would you have in `"300 mL"` of 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 `"5-mL"` sample, you should report the answer as

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