Calculate the number of electrons which will together weigh one gram

Solution:

(i) Mass of one electron $=9.10939 \times 10^{-31} \mathrm{~kg}$

$\therefore$ Number of electrons that weigh $9.10939 \times 10^{-31} \mathrm{~kg}=1$

$\therefore$ Number of electrons that will weigh $1 \mathrm{~g}\left(1 \times 10^{-3} \mathrm{~kg}\right)$

$=\frac{1}{9.10939 \times 10^{-31} \mathrm{~kg}} \times\left(1 \times 10^{-3} \mathrm{~kg}\right)$

$=0.1098 \times 10^{-3+31}$

$=0.1098 \times 10^{28}$

$=1.098 \times 10^{27}$

(ii) Mass of one electron $=9.10939 \times 10^{-31} \mathrm{~kg}$

Mass of one mole of electron $=\left(6.022 \times 10^{23}\right) \times\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)$

$=5.48 \times 10^{-7} \mathrm{~kg}$

Charge on one electron $=1.6022 \times 10^{-19}$ coulomb

Charge on one mole of electron $=\left(1.6022 \times 10^{-19} \mathrm{C}\right)\left(6.022 \times 10^{23}\right)$

$=9.65 \times 10^{4} \mathrm{C}$