A solute dissociate in solution according to reaction `2Ato5B`, If solute shows 30% dissociation then van't Haff factor will be:-

Options

A. 2.2
B. 1.45
C. 2.9
D. 1.9

Explanation

To find the Van't Hoff factor (i) for the dissociation of the solute according to the reaction \(2A \rightarrow 5B\) with 30% dissociation, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
The dissociation of the solute is given by the reaction \(2A \rightarrow 5B\). This means that 2 moles of A produce 5 moles of B.

2. **Define the Degree of Dissociation**:
Let \(\alpha\) be the degree of dissociation. Given that the solute shows 30% dissociation, we have:
\[ \alpha = 0.3 \]

3. **Initial Concentration**:
Assume we start with 2 moles of A (as per the stoichiometry of the reaction). At time \(t = 0\):
- Moles of A = 2
- Moles of B = 0

4. **Calculate Moles at Time \(t\)**:
After dissociation, the moles of A and B can be expressed as:
- Moles of A remaining = \(2 - 2\alpha = 2 - 2(0.3) = 2 - 0.6 = 1.4\)
- Moles of B formed = \(5\alpha = 5(0.3) = 1.5\)

5. **Total Moles After Dissociation**:
The total number of moles after dissociation is:
\[ \text{Total moles} = \text{Moles of A} + \text{Moles of B} = 1.4 + 1.5 = 2.9 \]

6. **Calculate the Van't Hoff Factor (i)**:
The Van't Hoff factor is calculated using the formula:
\[ i = \frac{\text{Total moles after dissociation}}{\text{Initial moles before dissociation}} = \frac{2.9}{2} \]
Thus,
\[ i = 1.45 \]

### Final Answer:
The Van't Hoff factor (i) is \(1.45\).

---