What fraction of the electrons would you have to remove from a 10.0 mg sphere of iron (Z=26,A=56)in order to make its charge 1.00 C? A)0.224% B)0.482% C)2.24 * 10^-4 D)4.82 * 10^-4 E)4.00 * 10^-7

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Quizplus · Accepted Answer

First, calculate the number of moles of iron in a 10.0 mg sphere using its molar mass (56 g/mol for iron): $10.0 \, \text{mg} = 0.01 \, \text{g}$, so $\frac{0.01 \, \text{g}}{56 \, \text{g/mol}} = 1.79 \times 10^{-4} \, \text{mol}$. The number of iron atoms is $1.79 \times 10^{-4} \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 1.08 \times 10^{20} \, \text{atoms}$. Since each iron atom contributes 26 electrons, the total number of electrons is $1.08 \times 10^{20} \times 26 = 2.80 \times 10^{21}$ electrons. To achieve a charge of +1.00 C (where the charge of one electron is $1.60 \times 10^{-19}$ C), the number of electrons needed to be removed is $\frac{1.00 \, \text{C}}{1.60 \times 10^{-19} \, \text{C/electron}} = 6.25 \times 10^{18}$ electrons. The fraction of electrons removed is $\frac{6.25 \times 10^{18}}{2.80 \times 10^{21}} = 0.00224$ or $0.224\%$.

What Fraction of the Electrons Would You Have to Remove