[JamesBond007a]

Here is the question:
A student determined the Ca^2+ ion content of a 100 mL sample of calcium-fortified orange juice. After adding 2.0 M HCl solution to the sample, the student filtered the juice to remove the pulp and other particulate matter. Then the student added Na2CO3 solution precipitate CaC03. The student recovered 0.2818 g of CaC03 from the resulting mixture.

Questions

A) Calculate the number of milligrams of Ca^2+ ion in the analyzed juice sample.

So I did (0.2818 g of CaCO3) * (40.08 g of Ca^+2 ion mol / 100.1 g of CaCO3 mol) = 1.128 x 10^-1 g, but in mg I would just * 1000 and get 112.8 mg

B) As prepared, the density of the juice was 0.958 g ml. Calculate the mass percent of Ca^2+ ion in the analyzed juice sample.

I did 1.128 x 10^-1 g/ 0.958 g mL and got  .1177 but * 100= 11.77%

C) According to the juice label, "each serving contains 20% of the USRDA of calcium," and one serving is defined as 6 oz or 178 mL. Does the student's analysis verify this claim? Briefly explain..

I think it's yes but I am not sure how to do the math or explain...

Do my answers look right in question A and B..

Thanks so much

[Borek]

A is OK, B is wrong - 0.958 is a mass of 1 mL, not of the analyzed sample.