Vant Hoff factor for a dilute aqueous solution of HCN is 1.00002. The percent degree of dissociation of the acid is :

Options

A. `2xx10^(-5)`
B. `2 xx 10^(-4)`
C. `2 xx 10^(-3)`
D. `2 xx 10^(-2)`

Answer

To find the percent degree of dissociation of HCN given the van't Hoff factor (i), we can follow these steps:

### Step 1: Understand the Van't Hoff Factor
The van't Hoff factor (i) is a measure of the number of particles into which a solute dissociates in solution. For HCN, which dissociates into H⁺ and CN⁻, the total number of ions (n) after dissociation is 2.

### Step 2: Use the Formula for Degree of Dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{i - 1}{n - 1} \]
Where:
- \(i\) is the van't Hoff factor
- \(n\) is the total number of ions formed after dissociation

### Step 3: Substitute the Known Values
From the problem:
- \(i = 1.0002\)
- \(n = 2\)

Substituting these values into the formula:
\[ \alpha = \frac{1.0002 - 1}{2 - 1} \]

### Step 4: Calculate α
Now, calculate α:
\[ \alpha = \frac{0.0002}{1} = 0.0002 \]

### Step 5: Convert α to Percentage
To find the percent degree of dissociation, multiply α by 100:
\[ \text{Percent Degree of Dissociation} = \alpha \times 100 = 0.0002 \times 100 = 0.02\% \]

### Final Answer
The percent degree of dissociation of HCN is **0.02%**.

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