What did I do wrong? Calculating the molarity of a solution

[Voltman]

Calculate the molarity of the solution:
0.64 g of H2SO4 dissolved in 500 mL

I have tried to do the above but got the wrong answer. Here is what I did:

Mr(H2SO4) = (1.00797*2) + (32.064*1) + (15.9994*3) = 82.07814
= 82.1 g/mol (rounded to least number significant figures)

n = m/M
n = 0.64 g/82.1 g/mol
= 0.007795371 mol
= 7.7*10^-3 mol (rounded to least number of significant figures)

c= n/V
c = 7.7*10^-3 mol/0.5 L
= 0.0154 M
= 2.0*10^-2 M (rounded to least number of significant figures)

However, the answer is supposed to be 0.013 M. What did I do wrong? Was it the methodology or did I round too much or was it something else? Any help on the above will be appreciated!...if possible please include all correct working out to above.

[Sonic]

Calculate the molarity of the solution:
0.64 g of H2SO4 dissolved in 500 mL

I have tried to do the above but got the wrong answer. Here is what I did:

Mr(H2SO4) = (1.00797*2) + (32.064*1) + (15.9994*3) = 82.07814
= 82.1 g/mol (rounded to least number significant figures)

n = m/M
n = 0.64 g/82.1 g/mol
= 0.007795371 mol
= 7.7*10^-3 mol (rounded to least number of significant figures)

c= n/V
c = 7.7*10^-3 mol/0.5 L
= 0.0154 M
= 2.0*10^-2 M (rounded to least number of significant figures)

However, the answer is supposed to be 0.013 M. What did I do wrong? Was it the methodology or did I round too much or was it something else? Any help on the above will be appreciated!...if possible please include all correct working out to above.

Your method is generally good, you have only calculated for 3 oxygen molecules in H2SO4 though when there are 4.

Mr(H2SO4): (1.00797*2) + (32.064*1) + (15.9994*4) = 98.07754

Your other main error is in rounding, never round off in the middle of a calculation, doing so will introduce rounding errors and these can have quite an effect on subsequent calculations. Instead round off at the end of a question - if the question is in parts round off at the end of each part but carry the original number to the next part (with rounding it is best to ask your tutor what you should do as this varies)

If you need to store numbers a lot of calculators have more memory slots at letters, on mine there is A to E in red which can be saved to

[SuBNoIze]

I realized you also had an error in calculating the first portion, which is what threw you off. Anyway, you wanted it worked out so:

Molarity = Moles of Solute/Liter of Solution

0.64 x 1/98.07754 = 0.0065254491497237797766950516907337

= 0.0065254491497237797766950516907337/0.500

= 0.013050898299447559553390103381467.