[j22rober]

Show mathematically how 0.2ml of a 0.02N sulphuric acid solution is equivalent to 0.02%?

If I convert my concentration to molarity (0.01M) and my volume to L (2x10^-4) then I can calculate the moles to give me 0.02....but this is not a percent. I am open to suggestions on this. Thanks.

[Dan]

If I convert my concentration to molarity (0.01M) and my volume to L (2x10^-4) then I can calculate the moles to give me 0.02

I agree with 0.01 M, but how you got 0.02 mol from there I don't know.

Remember, M = mol/L. A concentration of 0.01 M means 0.01 mol in one litre.

[AWK]

Volume in not important for concentration conversions. Since 0.01 N is a very diluted acid then you can use density of water for calculations.
0.01 N = 0.005 M => 0.005 mole /1000 g ~ 0.5 g /1000g = 0.05 %
Mole H₂SO₄ = 98 ~ 100

[Borek]

Show mathematically how 0.2ml of a 0.02N sulphuric acid solution is equivalent to 0.02%?

Question as worded makes no sense. 0.02% of what?

[j22rober]

Ok, I spoke to my teacher about this question today. Which was written as above in my assignment. Apparently I have to calculate the mass (g) of the sulfate ion (SO4) from this information. Where MW =  96.06g/mol, v=0.0002L, C=0.01mol/L. Combining C=N/V and N=m/MW gives: m=CxVxMW = 0.000192g which is ~ 0.0002g. Apparently I have a 1.0g (new info given to me today) sample containing  the 0.0002g. So if I have a 100g sample there is 0.02g in it. By weight percent (w/w)= (0.02g/100g)x 100% = 0.02%.

This is how I was supposed to solve it...who knew? Especially with the missing information.

[AWK]

_{MW =  96.06g/mol}
Why sulfate?
Your question concenrns sulfuric acid!