Chapter 16, Problem 53GP

To determine

To Find: Fraction of electrons lost by $3g$ of copper.

Answer to Problem 53GP

  $2.9\times {10}^{-10}$

Explanation of Solution

Given:

Mass of copper, (m) = $3g$

Charge on copper = $38\mu C$

Formula used:

Total charge = number of electrons $\times$ charge on single electron

  $\begin{array}{l}=Ne\\ =\left(\frac{m}{M}\times N_A\right)\times e\end{array}$

Here,

  $m=$ given mass of copper

  $M=$ molar mass of copper

  $N_A=$ Avogadro constant

  $e=$ charge on an electron

Calculation:

The fraction of the electrons lost is given by,

  $n_C=\frac{q}{\left(\frac{m_C}{M_{\text{and}}}\right)\text{Nne}}$

Calculate the fraction of the electrons lost as follows:

Substitute $38\times {10}^{-6}\text{C}$ for $q,3.0\text{g}$ for $m_C,\left(\frac{(1\text{mole})}{(63.5\text{g})}\right)$ for $M_{\text{atomic}},\left(\frac{6.02\times {10}^{23}\text{ atoms }}{\text{ mole }}\right)$ for $N,$

  $\left(\frac{29\text{e}}{\text{ atoms }}\right)$ for $n$ and $\left(\frac{1.602\times {10}^{-19}\text{C}}{1\text{e}}\right)$ for $N$ in the equation $\begin{array}{l}{n}_C=\frac{q}{\left(\frac{m_C}{M_{\text{atamic}}}\right)Nne}\\ n_C=\frac{38\times {10}^{-6}\text{C}}{(3.0\text{g})\frac{(1\text{mole})}{(63.5\text{g})}\left(\frac{6.02\times {10}^{23}\text{ atoms }}{\text{mole}}\right)\left(\frac{29\text{e}}{\text{ atoms }}\right)\left(\frac{1.602\times {10}^{-19}\text{C}}{1\text{e}}\right)}\\ =\frac{1}{3.5\times {10}^9}\\ =2.9\times {10}^{-10}\end{array}$

Therefore, the fraction of the electron lost is $2.9\times {10}^{-10}$ .

Conclusion:

  $2.9\times {10}^{-10}$part of electronic charge of $3g$ copper was lost in gaining $38\mu C$ .