I know how you obtained the wrong answer. You divided 0.06/105 and multiplied by 100 to get 0.0561 BUT you don't have the units right.
Atomic mass I = 126.9
molar mass I2 = 126.9 x 2 = 253.8
0.06 mol I2 = (253.8g/mol) x 0.06 mol = 15.22 g I2.
Total mass is 15.22 + 105 = ??
mass percent = (15.22/total mass)*100 = ?? about 13% or so. Check my thinking. Check my work.
I'll post the b part separately.
I have worked this problem out a number of ways, but every answer I come up with is incorrect. I have posted my last answers underneath the question. Please advise.
(a) What is the mass percentage of iodine (I2) in a solution containing 0.060 mol I2 in 105 g of CCl4?
My answer: 0.05711%
(b) Seawater contains 0.0079 g Sr2+ per kilogram of water. What is the concentration of Sr2+ measured in ppm?
My answer: 0.007900 ppm
2 answers
Do they, (they meaning the question makers) expect you to take into account the density of sea water since it isn't 1.00 g/mL. I will assume you are to consider density of sea water to be the same as fresh water. It won't make much difference in the answer.
Your answer you obtained by 0.0079 g Sr/1,000 g water and that is 0.0079 parts per thousand. Just set up a ratio of (0.0079 g Sr/1000 g H2O) = X g (Sr/1,000,000 g H2O) and solve for X. Check my thinking. Check my work.
Your answer you obtained by 0.0079 g Sr/1,000 g water and that is 0.0079 parts per thousand. Just set up a ratio of (0.0079 g Sr/1000 g H2O) = X g (Sr/1,000,000 g H2O) and solve for X. Check my thinking. Check my work.