On electrolysis of water, a total of 1 mole of gases is evolved. The amount of water decomposed is

Options

A. 1 mol
B. 2 mol
C. `(1)/(3)` mol
D. `(2)/(3)` mol

Explanation

To solve the problem of how much water is decomposed during the electrolysis of water when a total of 1 mole of gases is evolved, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The electrolysis of water can be represented by the following reaction:
\[ 2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \]
This indicates that 2 moles of water decompose to produce 2 moles of hydrogen gas and 1 mole of oxygen gas, resulting in a total of 3 moles of gas.

2. **Moles of Gas Produced**:
From the balanced equation, we see that for every 2 moles of water decomposed, 3 moles of gas are produced.

3. **Setting Up the Proportion**:
We know that the total amount of gas evolved is 1 mole. We can set up a proportion based on the stoichiometry of the reaction:
\[ \text{If } 3 \text{ moles of gas are produced from } 2 \text{ moles of water, then } 1 \text{ mole of gas will be produced from } x \text{ moles of water.} \]
This can be expressed mathematically as:
\[ \frac{3 \text{ moles of gas}}{2 \text{ moles of water}} = \frac{1 \text{ mole of gas}}{x \text{ moles of water}} \]

4. **Cross-Multiplying to Solve for x**:
Cross-multiplying gives us:
\[ 3x = 2 \implies x = \frac{2}{3} \text{ moles of water} \]

5. **Conclusion**:
Therefore, the amount of water decomposed when 1 mole of gases is evolved is:
\[ \frac{2}{3} \text{ moles of water} \]

### Final Answer:
The amount of water decomposed is \( \frac{2}{3} \) moles.

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