Question

Find the number of free electrons per unit volume of a metallic wire of density ${10}^4{\text{kg/m}}^3$ and atomic mass number $100$. The number of free electrons per atom is one.

• A. $6.022\times {10}^{23}$
• B. $6.022\times {10}^{25}$
• C. $6.022\times {10}^{28}$
• D. $6.022\times {10}^{30}$

Answer 1

The correct option is C $6.022\times {10}^{28}$

We know that, number of mole,

$n=\frac{\text{Mass}}{\text{Atomic mass}}=\frac{\text{Number of atom}}{\text{Avogadro Number}}$

$\Rightarrow \frac{{10}^4\times 1}{100\times {10}^{-3}}=\frac{x}{6.022\times {10}^{23}}[\because \rho =\frac{m}{V}]$

$\Rightarrow x=6.022\times {10}^{28}$

As number of free electrons per atom is $1$, so, number of free electrons will be $6.022\times {10}^{28}$.