Question

Excess electrons are placed on a small lead sphere with a mass of 7.85 gso that its net charge is −3.30×10⁻⁹ C .

Part A.

Find the number of excess electrons on the sphere.

ne =

Part B.

How many excess electrons are there per lead atom? The atomic number of lead is 82, and its molar mass is 207g/mol.

nlead =

Answer 1

Given that, net charge on the small lead sphere is Q = −3.30×10⁻⁹ C and mass of the lead sphere is m = 7.85 g

(A) Using the formula, charge Q = ne, where e is charge per electron which is -1.602 x 10^-19 C.

Therefore, the number of excess electrons on the sphere is

n_e = Q/e = ( −3.30×10⁻⁹ C) / (-1.602 x 10^-19 C) = 2.05 x 10¹⁰.

(B) Given that, molar mass is M = 207 g/mol and atomic number of Lead is 82

The number of moles of lead is equal to mass m / molar mass M = (7.85 g) / (207 g/mol) = 0.0379 mol.

Now, we can calculate the total number of lead atoms is n = N_A (m/M),

where N_A is Avogadro's number = 6.023 x 10²³ atoms/mol.

Therefore, n = ( 6.023 x 10²³ atoms/mol) (0.0379 mol) = 2.28 x 10²² atoms.

Finally, the number of excess electrons per lead atom is n_lead = number of excess electrons(n_e) / number of lead atoms (n)

then n_lead = (2.05 x 10¹⁰) / (2.28 x 10²² atom) = 9.0 x 10^-12 per atom.