a Calculate the number of electrons which will together

SOLUTION

Mass of one electron = $9.10939\times {10}^{–31}$ kg
$\therefore$ Number of electrons that weigh $9.10939\times {10}^{–31}$ kg = 1
$\therefore$ Number of electrons that will weigh 1 g (1 × ${10}^{–3}$ kg) $=\frac{1}{9.10939\times {10}^{-31}kg}\times (1\times {10}^{-31}kg)=0.1098\times {10}^{-3+31}=0.1098\times {10}^{28}=1.098\times {10}^{27}$

(b) Mass of one electron = $9.10939\times {10}^{–31}$ kg
Mass of one mole of electron = $6.022\times {10}^{23}\times 9.10939\times {10}^{–31}$ kg = $5.48\times {10}^{–7}$ kg
Charge on one electron = $1.6022\times {10}^{–19}$ coulomb
Charge on one mole of electron = $1.6022\times {10}^{–19}$ C $6.022\times {10}^{23}$  = $9.65\times {10}^4$C