Question

Excess electrons are placed on a small lead sphere with a mass of 7.70 g so that its net charge is −3.35 × 10^−9 C.A) Find the number of excess electrons on the sphere.
B) How many excess electrons are there per lead atom? The atomic number of lead is 82, and its atomic mass is 207 g/mol?

Answer 1

Answer:

a

$N = 2.094*10^{10} \ electrons$

b

$O = 9.33*10^{-13} \ electrons$

Explanation:

From the question we are told  that

   The  mass of the lead sphere is  $m = 7.70g = 0.0077 \ kg$

   The net charge is $Q_{net} = -3.35*10^{-9} \ C$

   The  atomic number is  $u = 82$

     The  molar mass is  $M = 207 \ g/mol$

Generally the excess number of  electron on the sphere is mathematically represented as

     $N = \frac{Q_{net}}{ e }$

Here e is the charge on the electron is $e = -1.60 *10^{-19} \ C$  

     So

    $N = \frac{-3.35 *10^{-19}}{ -1.60*10^{-19}}$

     $N = 2.094*10^{10} \ electrons$

Generally the number of atom present is mathematically represented as  

     $n = N_a * \frac{m}{ M}$

Here $N_a$ is the Avogadro's number with value  $N_a = 6.0*10^{23} \ atoms$

       $n = 6.03 *10^{23} * \frac{7.70}{ 207}$

      $n = 2.24 *10^{22} \ atoms$

Generally the electrons are there per lead atom is mathematically represented as

       $O = \frac{N}{n}$

=>   $O = \frac{2.24*10^{22}}{2.094*10^{10}}$

=>    $O = 9.33*10^{-13} \ electrons$