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
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
You can thus say that the
So, if you have
#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")))#