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 from the given Van't Hoff factor, we can follow these steps:

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

### Step 2: Use the Formula for Van't Hoff Factor
The formula for the Van't Hoff factor is given by:
\[ i = 1 + \alpha(n - 1) \]
where:
- \( i \) = Van't Hoff factor
- \( \alpha \) = degree of dissociation
- \( n \) = number of particles formed from one formula unit of solute

### Step 3: Substitute Known Values
From the question, we know:
- \( i = 1.00002 \)
- \( n = 2 \) (since HCN dissociates into 2 ions: H⁺ and CN⁻)

Substituting these values into the formula:
\[ 1.00002 = 1 + \alpha(2 - 1) \]

### Step 4: Simplify the Equation
Now simplify the equation:
\[ 1.00002 = 1 + \alpha(1) \]
\[ 1.00002 - 1 = \alpha \]
\[ \alpha = 0.00002 \]

### Step 5: Calculate Percent Degree of Dissociation
To find the percent degree of dissociation, we multiply \( \alpha \) by 100:
\[ \text{Percent degree of dissociation} = \alpha \times 100 = 0.00002 \times 100 = 0.002\% \]

### Final Answer
The percent degree of dissociation of HCN is:
\[ 0.002\% \]