# 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?

Dec 11, 2017

$\text{2 g}$

#### 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 $\text{100 mL}$ of the solution.

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

You can thus say that the $\text{5-mL}$ sample contained $\text{0.025 g}$ of potassium permanganate, the amount of solute present in the diluted solution.

So, if you have $\text{0.025 g}$ of solute in $\text{5 mL}$ of the solution, how many grams would you have in $\text{300 mL}$ 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 $\text{5-mL}$ sample, you should report the answer as

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{mass KMnO"_4 = "2 g}}}}$