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A $1.85 \mathrm{~g}$ sample of an arsenic-containing pesticide was chemically converted to $\mathrm{AsO}_{4}{ }^{3-}$ (atomic mass of $\mathrm{As}=$ 74.9) and titrated with $\mathrm{Pb}^{2+}$ to form $\mathrm{Pb}_{3}\left(\mathrm{AsO}_{4}\right)_{2}$ If $20 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{Pb}^{2+}$ is required to reach the equivalence point, the mass percentages of arsenic in the pesticide sample is closest to
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The correct answer is:
$5.4$
$3 \mathrm{pb}^{2+}+2 \mathrm{ASO}_{4}^{3-} \longrightarrow \mathrm{pb}_{3}\left(\mathrm{AsO}_{4}\right)_{2}$
$\mathrm{n}=\mathrm{M} \times \mathrm{V} \quad \mathrm{n}=\frac{2}{3} \times 2 \times 10^{-3}$
$=0.1 \times \frac{20}{1000}=0.00133$
$=2 \times 10^{-3}$
$\eta_{\mathrm{AS}}=\eta_{\mathrm{ASO}_{4}^{3}}=0.00133$
$\mathrm{~W}_{\mathrm{AS}}=0.00133 \times 74.9=0.0996$
$\%$ of $\mathrm{AS}=\frac{0.0996}{1.85} \times 100=5.4 \%$
$\mathrm{n}=\mathrm{M} \times \mathrm{V} \quad \mathrm{n}=\frac{2}{3} \times 2 \times 10^{-3}$
$=0.1 \times \frac{20}{1000}=0.00133$
$=2 \times 10^{-3}$
$\eta_{\mathrm{AS}}=\eta_{\mathrm{ASO}_{4}^{3}}=0.00133$
$\mathrm{~W}_{\mathrm{AS}}=0.00133 \times 74.9=0.0996$
$\%$ of $\mathrm{AS}=\frac{0.0996}{1.85} \times 100=5.4 \%$
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