If `alpha` is the degree of dissociation of `Na_(2)SO_(4)`, the van,t Hoff factor `(i)` used for calculating molaecular mass is:

Options

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

Explanation

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

### Step 1: Understand the dissociation of Na₂SO₄
Sodium sulfate dissociates in water according to the following reaction:
\[ \text{Na}_2\text{SO}_4 \rightarrow 2 \text{Na}^+ + \text{SO}_4^{2-} \]

### Step 2: Identify the number of particles produced
From the dissociation reaction, we see that one mole of Na₂SO₄ produces:
- 2 moles of Na⁺ ions
- 1 mole of SO₄²⁻ ions

Thus, one mole of Na₂SO₄ produces a total of:
\[ 2 + 1 = 3 \text{ moles of ions} \]

### Step 3: Consider the degree of dissociation (α)
Let’s denote the degree of dissociation as α. This means that if we start with 1 mole of Na₂SO₄, after dissociation:
- The amount of Na₂SO₄ that dissociates is α moles.
- The amount of Na₂SO₄ that remains undissociated is \( 1 - \alpha \) moles.

### Step 4: Calculate the moles of ions produced
After dissociation:
- The moles of Na⁺ produced = \( 2 \times \alpha \) (since each mole of Na₂SO₄ produces 2 moles of Na⁺)
- The moles of SO₄²⁻ produced = \( \alpha \)

Thus, the total moles of ions produced from the dissociation can be calculated as:
\[ \text{Total moles of ions} = \text{moles of Na}^+ + \text{moles of SO}_4^{2-} \]
\[ = 2\alpha + \alpha = 3\alpha \]

### Step 5: Calculate the van 't Hoff factor (i)
The van 't Hoff factor (i) is defined as the total number of particles in solution after dissociation compared to the number of formula units initially present. Therefore, we can express i as:
\[ i = \text{Total moles of ions} / \text{Initial moles of solute} \]
\[ i = \frac{(1 - \alpha) + 3\alpha}{1} \]
\[ = 1 - \alpha + 3\alpha \]
\[ = 1 + 2\alpha \]

### Conclusion
Thus, the van 't Hoff factor (i) for Na₂SO₄ in terms of the degree of dissociation (α) is:
\[ i = 1 + 2\alpha \]