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?

Question

Jeremiah Adams · Accepted Answer

#"2 g"#

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")))#

Peyton Adams · Answer

undefined To calculate the weight of potassium permanganate required, we first find the concentration of the stock solution using the given dilution information. Then, we use the formula for percent concentration to find the weight of potassium permanganate needed.

Concentration of stock solution = (Final concentration * Final volume) / Initial volume
= (0.01 * 250) / 5
= 0.5%

Now, we use the formula for percent concentration:
% w/v = (mass of solute / volume of solution) * 100

Rearranging for mass of solute:
mass of solute = (% w/v * volume of solution) / 100

mass of solute = (0.5 * 300) / 100
mass of solute = 1.5 grams

Therefore, 1.5 grams of potassium permanganate is required to produce 300 mL of solution.