Chemistry: A Molecular Approach (3rd Edition)

$9.4 \times 10^{13}$ excess electrons have been acquired. The collective mass is $8.5 \times 10^{-17}$ kg
Charge on one electron = $-1.6 \times 10^{-19}$ C Total accumulated charge = $-15 \times 10^{-6}$ C Hence, number of excess electrons = $\frac{(-15 \times 10^{-6})}{(-1.6 \times 10^{-19})}$ = $9.4 \times 10^{13}$ Mass of one electron = $9.10 \times 10^{-31}$ kg Hence, collective mass = $9.4 \times 10^{13}\times9.10 \times 10^{-31}$ kg = $8.5 \times 10^{-17}$ kg