Question

The total negative charge of all the electrons present in \(100\,g\) of water is

• A. \(8.90\times10^{8}\,C\)
• B. \(1.602\times10^{7}\,C\)
• C. \(5.360\times10^{8}\,C\)
• D. \(6.022\times10^{23}\,C\)

Hint:

A neutral molecule containing \(Z\) electrons contributes \(Z\) moles of electrons per mole of molecules. For water: \[ H_2O \rightarrow 10 \text{ electrons per molecule} \] Therefore, \[ 1\;mol\;H_2O \rightarrow 10\;mol\;e^- \] which simplifies charge calculations.

Answer 1

The Correct Option is C

Concept: Each water molecule contains \[\begin{aligned} 2(1)+8=10 \end{aligned}\] electrons. The total charge is calculated using \[\begin{aligned} Q=nF \end{aligned}\] where \(F=96500\,C\,mol^{-1}\).

Step 1: Calculate the moles of water. \[\begin{aligned} n(H_2O) &=\frac{100}{18}\\ &=5.556\,mol \end{aligned}\]

Step 2: Calculate moles of electrons. Each molecule of water contains \(10\) electrons. Hence, \[\begin{aligned} n(e^-) &=10\times5.556\\ &=55.56\,mol \end{aligned}\]

Step 3: Calculate the total electronic charge. \[\begin{aligned} Q &=nF\\ &=55.56\times96500\\ &=5.36\times10^{6}\,C \end{aligned}\]

Step 4: Match with the given options. \[ \begin{aligned} \text{Moles of water} &:\quad 5.556 \\ \text{Electrons per molecule} &:\quad 10 \\ \text{Moles of electrons} &:\quad 55.56 \\ \text{Total charge} &:\quad 5.36\times10^{6}\,\mathrm{C} \end{aligned} \] Thus, the intended answer among the given options is \[\begin{aligned} \boxed{5.360\times10^{6}\,C} \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.