Long ago. a workman at a dye factor fell into a vat containing hot, concentrated

Question

Long ago. a workman at a dye factor fell into a vat containing hot, concentrated H_2SO_4 and MNO_3. 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 the vat contained 8.00 times 10^3 L of liquid, and a 100, 0-mL sample was analyzed. If the man did fall into the vat. what is the expected concentration of phosphorus, in g/L, in 100.0 mL? The 100.0-mL sample was treated with moly date reagent that precipitated ammonium phosphor moly date. (NH_4)_3[P(Mo_12O_40)]*12H_2O. This substance was dried at 110degree C to remove wafers of hydration and heated to 400degree C until it reached the constant composition P_2O_5*24MoO_3, 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 P_2O_5*24MoO_3 (FM 3596.46) was produced. This blank determination gives the amount of phosphorus in the staring reagents, the P_2O_5*24MoO_3 that could have come from the dissolved man is therefore 0 3718 - 0 0331 = 0.3387 g. How much phosphorus was present in the 100.0-mL sample? Is this quantity consistent with a dissolved man?

Explanation / Answer

(a) weight of man = 70 kg = 70,000 g

1 g of human body has 6.3 mg of phosphorous

So, 70,000 g would have = 6.3 x 70,000 = 441,000 mg of phosphorous

441,000 mg phosphorous dissolved in 8 x 10^3 L vat solution

So, 0.1 L (100 ml) vat solution would have = 441,000 x 0.1/8 x 10^3

                                                                    = 5.5125 mg phosphorous in it.

(b) When the sample of acid was treated with molybdate and heated we get P2O5.12MoO3

mass of P2O5.12MoO3 in 100 ml sample = 0.3387 g

So,

mass of phosphorous in 100 ml sample = 0.3387 x 2 x 31/3596.46

                                                                = 5.84 mg

(c) the two values from (a) and (b) are close to one another, thus this quantity is consitent with dissolved man.

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