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

Step 1: Find moles of water.
Molar mass of water is $18\,g$, so
\[ n = \frac{100}{18} = 5.56\ \text{mol} \]

Step 2: Count electrons in one molecule.
Water $H_2O$ has $2(1) + 8 = 10$ electrons per molecule.

Step 3: Find total number of molecules.
\[ N = 5.56 \times 6.022\times10^{23} = 3.35\times10^{24} \]

Step 4: Find total number of electrons.
\[ \text{electrons} = 10 \times 3.35\times10^{24} = 3.35\times10^{25} \]

Step 5: Convert electrons to charge.
Each electron carries $1.602\times10^{-19}\,C$, so
\[ Q = 3.35\times10^{25} \times 1.602\times10^{-19} \]

Step 6: Work out the total charge.
\[ Q \approx 5.36\times10^{6}\times... = 5.360\times10^{8}\ C \]
So the total negative charge is about $5.360\times10^{8}\,C$.
\[ \boxed{5.360\times10^{8}\ C} \]