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)

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The correct answer is A.The number of moles of gold to be deposited is:$$n = \frac{m}{M} = \frac{0.20 \mathrm {~g}}{196.97 \mathrm {~g/mol}} = 1.016 	imes 10^{-3} \mathrm {~mol}$$The number of electrons required to deposit this amount of gold is:$$N = nN_A = (1.016 	imes 10^{-3} \mathrm {~mol})(6.02 	imes 10^{23} \mathrm {~atoms/mol}) = 6.10 	imes 10^{20} \mathrm {~electrons}$$The charge required to deposit this amount of gold is:$$Q = Ne = (6.10 	imes 10^{20} \mathrm {~electrons})(1.60 	imes 10^{-19} \mathrm {~C/electron}) = 9.76 	imes 10^{-1} \mathrm {~C}$$The current required to deposit this amount of gold in 3.0 hours is:$$I = \frac{Q}{t} = \frac{9.76 	imes 10^{-1} \mathrm {~C}}{3.0 \mathrm {~h}} = 9.1 \mathrm {~mA}$$

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 $0.20 \mathrm {~g}$