Question

Calculate number of electrons present in $500c{m}^3$ volume of water.

• A. $1.67\times {10}^{20}$
• B. $1.67\times {10}^{26}$
• C. $1.67\times {10}^{23}$
• D. $3.2\times {10}^{26}$

Answer 1

The correct option is B $1.67\times {10}^{26}$

$\begin{array}{c}\text{Given}\\ \text{of electron in one molecule of}{H}_{2}O=2+8\\ =10\end{array}$

$\begin{array}{c}\text{we know}\\ \text{density of watere}1g/{cm}^3\\ \text{given volume}=500{cm}^3\\ \text{No. of mole of water}=\left(\frac{500}{18}\right)\end{array}$

$\begin{array}{c}\text{Since}1\text{mole contain}6.02\times {10}^{23}\text{molecule of}{H}_{2}O\\ \text{and}1H_{2}O\text{contain}10e^{-}\end{array}$

$\begin{array}{c}\begin{array}{cc}\text{total}{e}^{-}\text{contain}& =\frac{500}{18}\times 6.022\times {10}^{23}\times 10\\ & =1.67\times {10}^{26}\end{array}\end{array}$