The no.of `H_(3)O^(+)` ions present in 10 ml of water at `25^(@)`C is

Options

A. `6.023 xx 10^(-14) `
B. `6.023 xx 10^(14 )`
C. `6.023 xx 10^(-19 )`
D. `6.023 xx 10^(19)`

Answer

To find the number of \( H_3O^+ \) ions present in 10 ml of water at \( 25^\circ C \), we can follow these steps:

### Step 1: Determine the concentration of \( H_3O^+ \) ions in water at \( 25^\circ C \)
At \( 25^\circ C \), the concentration of \( H_3O^+ \) ions in pure water is \( 1 \times 10^{-7} \) moles per liter (M).

### Step 2: Convert the volume of water from ml to liters
We have 10 ml of water. To convert this to liters:
\[ 10 \text{ ml} = \frac{10}{1000} \text{ L} = 0.01 \text{ L} \]

### Step 3: Calculate the number of moles of \( H_3O^+ \) in 10 ml of water
Using the concentration and the volume, we can find the number of moles of \( H_3O^+ \):
\[ \text{Number of moles} = \text{Concentration} \times \text{Volume} = (1 \times 10^{-7} \text{ moles/L}) \times (0.01 \text{ L}) = 1 \times 10^{-9} \text{ moles} \]

### Step 4: Calculate the number of \( H_3O^+ \) ions using Avogadro's number
Avogadro's number (\( N_A \)) is approximately \( 6.022 \times 10^{23} \) ions/mole. To find the total number of \( H_3O^+ \) ions:
\[ \text{Number of ions} = \text{Number of moles} \times N_A = (1 \times 10^{-9} \text{ moles}) \times (6.022 \times 10^{23} \text{ ions/mole}) \]
\[ = 6.022 \times 10^{14} \text{ ions} \]

### Final Answer
The number of \( H_3O^+ \) ions present in 10 ml of water at \( 25^\circ C \) is \( 6.022 \times 10^{14} \) ions.

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