One litre of water contains `10^(-7)` mole `H^(+)` ions. Degree of ionisation of water is:

Options

A. `1.8xx10^(-7)%`
B. `1.8xx10^(-9)%`
C. `3.6xx10^(-7)%`
D. `3.6xx10^(-9)%`

Explanation

To find the degree of ionization of water, we can follow these steps:

### Step 1: Understand the concept of degree of ionization
The degree of ionization (α) is defined as the fraction of the total number of moles that have dissociated into ions. It can be expressed mathematically as:
\[ \alpha = \frac{\text{Number of moles dissociated}}{\text{Total number of moles present}} \]

### Step 2: Determine the number of moles of \( H^+ \) ions
According to the question, one litre of water contains \( 10^{-7} \) moles of \( H^+ \) ions. Therefore, the number of moles dissociated is:
\[ \text{Number of moles dissociated} = 10^{-7} \text{ moles} \]

### Step 3: Calculate the total number of moles of water in one litre
To find the total number of moles of water in one litre, we can use the molar mass of water, which is approximately 18 g/mol. Since 1 litre of water has a mass of about 1000 g, we can calculate the total moles of water as follows:
\[ \text{Total moles of water} = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ moles} \]

### Step 4: Substitute the values into the formula for degree of ionization
Now, we can substitute the values we have into the formula for degree of ionization:
\[ \alpha = \frac{10^{-7}}{55.56} \]

### Step 5: Calculate the degree of ionization
Now, we perform the calculation:
\[ \alpha \approx 1.8 \times 10^{-9} \]

### Step 6: Convert to percentage (if required)
To express the degree of ionization as a percentage, we can multiply by 100:
\[ \alpha \text{ (in percentage)} = 1.8 \times 10^{-9} \times 100 = 1.8 \times 10^{-7} \% \]

Thus, the degree of ionization of water is approximately \( 1.8 \times 10^{-9} \) or \( 1.8 \times 10^{-7} \% \).

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