Question

The total number of electrons in $1.6\text{g}$ of $C{H}_4$ to that in $1.8\text{g}$ of $H_{2}O$ is:

• A. Double
• B. The same
• C. Triple
• D. One fourth

Answer 1

The correct option is B The same

Number of $e^{-}$ in $1.6\text{g}$ of $C{H}_4$$\begin{array}{c}=\frac{1.6}{16}\times 10\times N_A=N_A\\ [\text{Total number of}{e}^{-}\text{in}C{H}_4=10]\end{array}$

Number of $e^{-}$ in $1.8\text{g}$ of $H_{2}O$$\begin{array}{c}=\frac{1.8}{18}\times 10\times N_A=N_A\\ [\text{Total number of}{e}^{-}\text{in}{H}_{2}O=10]\end{array}$