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

Question

Quizplus · Accepted Answer

First, calculate the number of moles of gold to be deposited: $0.20 \, \text{g} / 196.97 \, \text{g/mol} = 1.015 \times 10^{-3} \, \text{mol}$. Then, find the total charge needed: $1.015 \times 10^{-3} \, \text{mol} \times 6.02 \times 10^{23} \, \text{atoms/mol} \times 1.60 \times 10^{-19} \, \text{C} = 9.744 \times 10^{4} \, \text{C}$. Finally, calculate the current required for 3.0 hours ($3.0 \times 3600 \, \text{s}$): $9.744 \times 10^{4} \, \text{C} / (3.0 \times 3600 \, \text{s}) = 9.04 \, \text{mA}$, which rounds to 9.1 mA.

A Jeweler Needs to Electroplate Gold,having an Atomic Mass of 196.97