The van't Hoff factor of a 0.005 M aqueous solution of KCl is 1.95. The degree of ionisation of KCl is :

Options

A. 0.95
B. 0.97
C. 0.94
D. 0.96

Explanation

To find the degree of ionization (α) of KCl in a 0.005 M aqueous solution where the van't Hoff factor (i) is given as 1.95, we can follow these steps:

### Step 1: Understand the dissociation of KCl
KCl dissociates in water into two ions:
\[ \text{KCl} \rightarrow \text{K}^+ + \text{Cl}^- \]
Thus, for KCl, the number of ions (n) produced upon dissociation is:
\[ n = 2 \]

### Step 2: Use the van't Hoff factor equation
The van't Hoff factor (i) is related to the degree of ionization (α) by the formula:
\[ i = 1 + \alpha (n - 1) \]
Where:
- \( i \) = van't Hoff factor
- \( \alpha \) = degree of ionization
- \( n \) = number of ions produced

### Step 3: Substitute the known values
We know:
- \( i = 1.95 \)
- \( n = 2 \)

Substituting these values into the equation:
\[ 1.95 = 1 + \alpha (2 - 1) \]

### Step 4: Simplify the equation
This simplifies to:
\[ 1.95 = 1 + \alpha \]
Now, isolate α:
\[ \alpha = 1.95 - 1 \]
\[ \alpha = 0.95 \]

### Step 5: Conclusion
The degree of ionization (α) of KCl in the solution is:
\[ \alpha = 0.95 \]

### Final Answer
The degree of ionization of KCl is 0.95 or 95%.

---