A jeweler needs to electroplate gold, having an atomic mass of 196.97 g/mol, onto a bracelet. He knows that the charge carriers in the ionic solution are singly-ionized gold ions, Au⁺, and has calculated that he must deposit $0.20 \mathrm {~g}$ of gold to reach the necessary thickness. How much current does he need to plate the bracelet in 3.0 hours? (e = 1.60 × 10^-19 C, N_A = 6.02 × 10²³ atoms/mol) A) 9.1 mA B) 540 mA C) 33 A D) 1800 mA

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Quizplus · Accepted Answer

To find the current, first calculate the number of moles of gold: $0.20 \, \text{g} / 196.97 \, \text{g/mol} = 0.001015 \, \text{mol}$. The number of gold ions is $0.001015 \, \text{mol} \times 6.02 \times 10^{23} \, \text{ions/mol} = 6.11 \times 10^{20} \, \text{ions}$. The total charge is $6.11 \times 10^{20} \, \text{ions} \times 1.60 \times 10^{-19} \, \text{C/ion} = 97.76 \, \text{C}$. Current is charge/time, so $97.76 \, \text{C} / (3 \times 3600 \, \text{s}) = 0.0091 \, \text{A} = 9.1 \, \text{mA}$.

A Jeweler Needs to Electroplate Gold, Having an Atomic Mass 0.20 g0.20 \mathrm {~g}0.20 g

A Jeweler Needs to Electroplate Gold, Having an Atomic Mass $0.20 \mathrm {~g}$