Question

Avagadro's number of carbon atoms have a mass of 12.0 g. Two identical graphite spheres each have a mass of m = 4.21 g. If we could transfer one electron from each atom in the first sphere to an atom in the second sphere, what would be the magnitude of the attractive force between the spheres if they are separated by 1.00 m?

Answer 1

Given that Mass of graphite sphere = 4.21 gm

Molecular weight of carbon = 12.0 gm/mol, So

moles = Mass/Molecular weight

moles = 4.21/12 = 0.35083 moles

Number of atoms of carbon in each sphere will be:

N = n*Na

Na = Avagadro's number = 6.023*10^23

N = 0.35083*6.023*10^23

N = 2.113*10^23 atoms

Now given that one electron from each atoms is transferred to another sphere, So total electrons transferred from 1st sphere to 2nd sphere

Total charge on 2nd sphere = Q2 = n*e

n = number of electrons transferred = 2.113*10^23 electrons

Charge on each electron = -1.6*10^-19 C

Q2 = 2.113*10^23*(-1.6*10^-19) = -33808 C

So charge on 1st sphere will be:

Q1 = |Q2| = |-33808 C| = 33808 C

Now electrostatic force is given by:

Fe = k*Q1*Q2/r^2

r = distance between each sphere = 1.00 m

k = 9*10^9

So, since Q1 and Q2 have opposite sign, So force will be attractive, Now magnitude of force will be

Fe = 9*10^9*33808^2/1.00^2

Fe = 1.03*10^19 N

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