If `alpha` is the degree of dissociation of `na_(2)SO_(4)` the vant of Hoff's factor (i) used for calculating the molecular mass is-

Options

A. `1-alpha`
B. `1+alpha`
C. `1-2alpha`
D. `1+2alpha`

Explanation

To find the van 't Hoff factor (i) for sodium sulfate (Na₂SO₄) in terms of its degree of dissociation (α), we can follow these steps:

### Step 1: Understand the dissociation of Na₂SO₄
Sodium sulfate dissociates in solution as follows:
\[ \text{Na}_2\text{SO}_4 \rightarrow 2 \text{Na}^+ + \text{SO}_4^{2-} \]

This means that one formula unit of Na₂SO₄ produces a total of 3 ions in solution: 2 sodium ions (Na⁺) and 1 sulfate ion (SO₄²⁻).

### Step 2: Define the degree of dissociation (α)
Let α be the degree of dissociation of Na₂SO₄. This means that a fraction α of the Na₂SO₄ dissociates into ions. If we start with 1 mole of Na₂SO₄, then:
- The amount that dissociates = α moles
- The amount that remains undissociated = (1 - α) moles

### Step 3: Calculate the total number of ions produced
When Na₂SO₄ dissociates:
- From α moles of Na₂SO₄, we get:
- 2α moles of Na⁺ ions
- 1α moles of SO₄²⁻ ions

Thus, the total number of moles of ions produced when Na₂SO₄ dissociates is:
\[ \text{Total ions} = 2\alpha + \alpha = 3\alpha \]

### Step 4: Calculate the van 't Hoff factor (i)
The van 't Hoff factor (i) is defined as the total number of particles (ions) in solution after dissociation. For Na₂SO₄, the total number of particles can be expressed as:
\[ i = 1 + (n - 1) \cdot \alpha \]
where n is the total number of ions produced from one formula unit of Na₂SO₄.

Since we have 3 ions (2 Na⁺ and 1 SO₄²⁻), we can substitute n = 3 into the equation:
\[ i = 1 + (3 - 1) \cdot \alpha \]
\[ i = 1 + 2\alpha \]

### Final Answer
Thus, the van 't Hoff factor (i) used for calculating the molecular mass of Na₂SO₄ is:
\[ i = 1 + 2\alpha \]

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