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 step by step, we need to find the number of `CN^(-)` ions in a 250 ml solution of 1.0 M HCN that is 1% ionized.

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

### Step 2: Calculate the number of moles of HCN
Since the concentration of HCN is 1.0 M, the number of moles in 250 ml (which is 0.250 L) can be calculated using the formula:
\[ \text{Number of moles of HCN} = \text{Concentration} \times \text{Volume} = 1.0 \, \text{mol/L} \times 0.250 \, \text{L} = 0.250 \, \text{mol} \]

### Step 3: Calculate the number of moles of `CN^(-)` ions
Since 1% of HCN ionizes, the number of moles of `CN^(-)` ions produced is:
\[ \text{Number of moles of } CN^{-} = \text{Number of moles of HCN} \times \alpha = 0.250 \, \text{mol} \times 0.01 = 0.0025 \, \text{mol} \]

### Step 4: Convert moles of `CN^(-)` ions to number of ions
To find the total number of `CN^(-)` ions, we multiply the number of moles by Avogadro's number (\(6.022 \times 10^{23} \, \text{ions/mol}\)):
\[ \text{Number of } CN^{-} \text{ ions} = 0.0025 \, \text{mol} \times 6.022 \times 10^{23} \, \text{ions/mol} \]
\[ = 1.5075 \times 10^{21} \, \text{ions} \]

### Step 5: Rounding the result
Rounding this to two significant figures, we get:
\[ \text{Number of } CN^{-} \text{ ions} \approx 1.5 \times 10^{21} \, \text{ions} \]

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

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