Calculate the number of electrons in a small, electrically neutral silver pin that has a mass of $$10.0g$$. Silver has $$47$$ electrons per atom, and its molar mass is $$107.87g$$ $${mol}^{-1}$$.
No. of atoms in $$10g$$ of silver,
$$n=\cfrac { 6.023\times { 10 }^{ 23 }\times 10 }{ 107.87 } =5.58\times { 10 }^{ 22 }\quad $$
No. of electrons $$=47n=2.62\times { 10 }^{ 24 }$$