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
Dec 11, 2017

Answer:

#"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")))#