Question #388f4

1 Answer
Jul 16, 2015

That depends on what exactly you're dealing with.


If I understand your question correctly, you're dealing with a solution that is has a percent concentration of 3.48%.

Now, let's assume that this percent concentration is weight by volume, w/v.

In simple terms, a 3.48% w/v solution of something contains 3.48 g of that something in every 100 mL of solution. If that is the case, 10 mL of this solution would contain 10 times fewer grams

#10cancel("mL") * "3.48 g"/(100cancel("mL")) = "0.348 g"#

If you were to take a 1000-mL sample of this olution, than it would contain

#1000cancel("mL") * "3.48 g"/(100cancel("mL")) = "34.8 g"#

The volume is 100 times bigger, the solution will contain 100 times more solute.

Now, let's assume that you want to perform a dilution of the 10-mL solution and increase its volume to 1000-mL.

In this case, you would add enough water to the first solution to get the total volume to 1000 mL - you'd add 990 mL of water, to be precise.

The solution would still contain 0.348 g of something, since you only added water, but its weight by volume percent concentration would be

#"0.348 g"/"1000 mL" * 100 = "0.0348%"#

This new solution wil contain 0.0348 g of something for every 100 g of solution.