a) 70,000 g man * (6.3 mg P/g man) (100 mL/ 8.00x10^3x10^3 mL) = 5.5125 mg P
b) 0.3718g - 0.0331g = 0.3387g P2O5.24MoO3
0.3387g P2O5.24MoO3 (1 mol/3596.46g P2O5.24MoO3) (2 mol P/mol P2O5.24MoO3) (30.97 g/mol P) = 5.83326 x 10^-3 g
= 5.833 mg P. Yes, the man died in the vat. :(
Man in the vat problem. Long ago, a workman at a dye factory fell into a vat containing a hot concentrated mixture of sulfuric and nitric acids. He dissolved completely! Because nobody witnessed the accident, it was necessary to prove that he fell in so that the man’s wife could collect his insurance money. The man weighed 70 kg, and a human body contains ~6.3 parts per thousand (mg/g) phosphorus. The acid in the vat was analyzed for phosphorus to see whether it contained a dissolved human.
(a) The vat contained 8.00 103 L of liquid, and a 100.0-mL sample was analyzed. If the man did fall into the vat, what is the expected quantity of phosphorus in 100.0 mL?
(b) The 100.0-mL sample was treated with a molybdate reagent that caused ammonium phosphomolybdate, (NH4)3[P(Mo12O40)]12H2O, to precipitate. This substance was dried at 110°C to remove waters of hydration and heated to 400°C until it reached a constant composition corresponding to the formula P2O524MoO3 (FM 3596.46), which weighed 0.3718 g. When a fresh mixture of the same acids (not from the vat) was treated in the same manner, 0.0331 g of P2O524MoO3 was produced. This blank determination gives the amount of phosphorus in the starting reagents. How much phosphorus was present in the 100.0-mL sample? Is this quantity consistent with a dissolved man?
1 answer
1 answer