Man in the vat problem.11 Long ago, a workman at a dye factory fell into a vat
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(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 precipitated ammonium phosphomolybdate, (NH4)3[P(Mo12O4O)] 12H2O. This substance was dried at 110°C to remove waters of hydration and heated to 400°C until it reached the constant composition P2O5 24MoO3, which weighed 0.3718 g. When a fresh mixture of the same acids (not from the vat) was treated in the same manner, 0.033 1 g of P2O5 24MoO3 (FM 3 596.46) was produced. This blank determination gives the amount of phosphorus in the starting reagents. The P2O5 24MoO3 that could have come from the dissolved man is therefore 0.371 8 - 0.0331 - 0.3387 g. How much phosphorus was present in the 100.0-mL sample? Is this quantity consistent with a dissolved man?
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