Question

A 50-g ball of copper has a net charge of 1.5 μC. What fraction of the copper's electrons have been removed? (Each neutral copper atom has 29 protons and 29 electrons, and copper has an atomic mass of 63.5 a.m.u.)

Answer 1

`Fraction\ of\ electrons\ lost=6.82* 10^(-18)`

Step-by-step explanation:

Mass of copper = 50 g

Molar mass of copper = 63.5 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (50\ g)/(63.5\ g/mol)`

`Moles= 0.7874\ mol`

Also,

1 atom of neutral copper contains 29 electrons

1 mole of neutral copper contains
`29* 6.023* 10^(23)` electrons

0.7874 moles of neutral copper contains
`29* 6.023* 10^(23)* 0.7874` electrons

0.7874 moles of neutral copper contains
`1.3753* 10^(25)` electrons.

Given, Charge = 1.5 μC

1 μ = 10⁻⁶ C

So, Charge on the copper =
`1.5* 10^(-6)\ C`

Charge on 1 electron =
`1.6* 10^(-19)\ C`

Thus, number of electrons = Total charge / Charge on one electron

Thus,

`n=\frac {1.5* 10^(-6)}{1.6* 10^(-19)}=9.375* 10^(12)`

Fraction :

`Fraction\ of\ electrons\ lost=\frac {9.375* 10^(12)}{1.3753* 10^(25)}=6.82* 10^(-18)`