CHEM: I am not sure what the question means..any help/hints will do...

You weight out 0.2023 g of primary Standard Na2CO3, 99.85% pure and it took 41.0 ml of 0.1M HCl to completely protonate the Na2CO3 in 0.2023g sample of Standard.

Net eq:
Na2CO3 + 2HCl ---> 2NaCl + H2O(l) + CO2 (g)

1. What is the mass of Na2CO3 in your 0.2023 g sample of standard?

I am guessing that the purity be multiplied by the given sample weight...?

Thanks for help. I really need it

Yes, you are correct, when you are accurately weighing out your standards you must consider the purity of the standard. It would also be good to know what the remaining 0.15% of the sample is.

mass Na2CO3 = mass primary standard x 0.9985 = (0.2023 g)(0.9985) = 0.2020 g

You will need to use this mass to calculate moles of Na2CO3 and after the titration moles of HCl.

Hope this helps you. JIL HIR