1.0 M HCN solution is 1.0 percent ionised. The number of `CN^(-)` ions in 250 ml of the solution is :

Options

A. `1.5 xx 10^(21)`
B. `6 xx 10^(23)`
C. `5 xx 10^(-3)`
D. `3.01 xx 10^(21)`

Answer

To solve the problem of finding the number of `CN^(-)` ions in a 250 ml solution of 1.0 M HCN that is 1% ionized, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Ionization Percentage**:
Given that the 1.0 M HCN solution is 1% ionized, we can express this as:
\[ \text{Ionization fraction} (\alpha) = \frac{1}{100} = 0.01 \]

2. **Calculate the Concentration of `CN^(-)` Ions**:
The concentration of `CN^(-)` ions produced from the ionization of HCN can be calculated using the formula:
\[ [CN^-] = C \times \alpha \]
where \( C \) is the initial concentration of HCN (1.0 M). Thus,
\[ [CN^-] = 1.0 \, \text{M} \times 0.01 = 0.01 \, \text{M} \]

3. **Convert Volume from ml to L**:
The volume of the solution is given as 250 ml. We need to convert this to liters:
\[ \text{Volume in liters} = \frac{250 \, \text{ml}}{1000} = 0.250 \, \text{L} \]

4. **Calculate the Number of Moles of `CN^(-)` Ions**:
The number of moles of `CN^(-)` ions can be calculated using the formula:
\[ \text{Number of moles} = [CN^-] \times \text{Volume} \]
Substituting the values we have:
\[ \text{Number of moles} = 0.01 \, \text{mol/L} \times 0.250 \, \text{L} = 0.0025 \, \text{mol} \]

5. **Calculate the Number of `CN^(-)` Ions**:
To find the total number of `CN^(-)` ions, we use Avogadro's number (\( N_A = 6.022 \times 10^{23} \) entities/mol):
\[ \text{Number of ions} = \text{Number of moles} \times N_A \]
Thus,
\[ \text{Number of ions} = 0.0025 \, \text{mol} \times 6.022 \times 10^{23} \, \text{ions/mol} \]
\[ \text{Number of ions} \approx 1.507 \times 10^{21} \, \text{ions} \]

### Final Answer:
The number of `CN^(-)` ions in 250 ml of the solution is approximately \( 1.5 \times 10^{21} \) ions.

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