Man in the vat problem.15 Long ago, a workman at a dye factory fell into a vat containing hot, concentrated 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$\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{L}$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,$\mathbf{\left(}{\mathbf{NH}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{3}}\mathbf{\right[}\mathbf{P}\mathbf{\left(}{\mathbf{Mo}}_{\mathbf{1}}\mathbf{2}{\mathbf{O}}_{\mathbf{4}}\mathbf{0}\mathbf{\right)}\mathbf{]}\mathbf{}\mathbf{12}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$This substance was dried at$\mathbf{110}\mathbf{°}\mathbf{C}$to remove waters of hydration and heated to$\mathbf{400}\mathbf{°}\mathbf{C}$until it reached the constant composition${\mathbf{P}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{5}}\mathbf{\times }\mathbf{24}{\mathbf{MoO}}_{\mathbf{3}}$, which weighed 0.371 8 g. When a fresh mixture of the same acids (not from the vat) was treated in the same manner, 0.033 1 g of${\mathbf{P}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{5}}\mathbf{\times }\mathbf{24}{\mathbf{MoO}}_{\mathbf{3}}\mathbf{(}\mathbf{FM}\mathbf{3596}\mathbf{.}\mathbf{46}\mathbf{)}$was produced. This blank determination gives the amount of phosphorus in the starting reagents. The${\mathbf{P}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{5}}\mathbf{\times }\mathbf{24}{\mathbf{MoO}}_{\mathbf{3}}$that could have come from the dissolved man is therefore$\mathbf{0}\mathbf{.}\mathbf{3718}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{0331}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3387}\mathbf{g}$.How much phosphorus was present in the 100.0-mL sample? Is this quantity consistent with a dissolved man?

Step by step solution

01

Calculate the quantity of phosphorus:

Calculate the quantity of phosphorus in$8\times {10}^3\mathrm{L}$,

$$\begin{gathered} m_1=m\left(man\right)\times \rho (Pperkg) \\ {m}_1=70kg\times 6.3g/kg \\ {m}_1=441g \end{gathered}$$

In 100 mL,

$$\begin{gathered} {\mathrm{m}}_2={\mathrm{m}}_2/8\times {10}^3\mathrm{L} \\ {\mathrm{m}}_2=441\mathrm{g}/(8\times {10}^3\mathrm{L}) \\ {\mathrm{m}}_2=0.0551\mathrm{g}/\mathrm{L} \\ =5.51\mathrm{mg}\mathrm{in}100\mathrm{mL} \end{gathered}$$

The quantity of phosphorus is 5.51mg.

02

Calculate mass of phosphorus in 100mL:

$$\begin{gathered} x\left(P\right)=\frac{2M\left(P\right)}{M(P_2{O}_5\times 24Mo{O}_3)} \\ x\left(P\right)=\frac{2\times 30.974g/mol}{3596.46g/mol} \\ x\left(P\right)=1.722\% \end{gathered}$$

Mass of phosphorus in

$$\begin{gathered} m\left(P\right)=x\left(P\right)\times 0.3387g \\ m\left(P\right)=0.01722\times 0.3387g \\ m\left(P\right)=5.834g \end{gathered}$$

The mass of phosphorus in 100mL is 5.834g

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