Answer
(a) $7.25 \times 10^{24}$ electrons.
(b) $5.27 \times 10^{15}$ electrons.
(c) $7.27 \times 10^{-10}$ electrons.
Work Step by Step
For (a):
The atomic mass of alumnium = 26.982 g/mol = 0.026982 kg/mol.
Each atom of Al has 13 electrons.
We know that 1 mol of Al contains $N_a = 6.022 \times 10^{23}$ atoms, so the number of atoms in a 0.0250 kg sphere = $(N_a) \frac{(0.0250)}{(0.026982)}$
The number of electrons $n$, then = $(13)(N_a) \frac{(0.0250)}{(0.026982)} = 7.25 \times 10^{24}$
For (b):
From Coulomb's law, $F = k\frac{q_1 q_2} {r^2}$. Let the charge of each sphere be $q$.
$\implies q = \sqrt{4 \pi \epsilon_0 Fr^2} = \sqrt{4 \pi \epsilon_0 (10^4)(0.8)^2} = 8.43 \times 10^{-4}$
Thus the number of electrons $n'$ removed from one sphere = $q/e = \frac{8.43 \times 10^{-4}}{1.6 \times 10^ {-19} }= 5.27 \times 10^{15} $
For (c):
$\frac{n'}{n} = \frac {5.27 \times 10^{15}}{7.25 \times 10^{24}} = 7.27 \times 10^{-10}$
