Assuming each salt to be `90%` dissociated which of the following will have the highest osmotic pressure?

Options

A. Decimolar `Al_(2)(SO_(4))_(3)`
B. `Decimolar `BaCl_(2)`
C. `Decimolar `Na_(2)SO_(4)`
D. `A solution obtained by mixing equal volumes of (B) and (C) and fitering.

Explanation

To determine which salt will have the highest osmotic pressure when each salt is assumed to be 90% dissociated, we can follow these steps:

### Step 1: Understand the formula for osmotic pressure
The osmotic pressure (π) can be calculated using the formula:
\[ \pi = iCRT \]
where:
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(C\) = molarity of the solution
- \(R\) = universal gas constant
- \(T\) = temperature in Kelvin

### Step 2: Identify the dissociation of each salt
We need to find the van 't Hoff factor \(i\) for each salt, which represents the number of particles produced when the salt dissociates in solution.

1. **For Al₂(SO₄)₃:**
- Dissociation:
\[ \text{Al}_2(\text{SO}_4)_3 \rightarrow 2 \text{Al}^{3+} + 3 \text{SO}_4^{2-} \]
- Total ions produced: \(2 + 3 = 5\)
- \(i = 5\)

2. **For BaCl₂:**
- Dissociation:
\[ \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2 \text{Cl}^- \]
- Total ions produced: \(1 + 2 = 3\)
- \(i = 3\)

3. **For Na₂SO₄:**
- Dissociation:
\[ \text{Na}_2\text{SO}_4 \rightarrow 2 \text{Na}^+ + \text{SO}_4^{2-} \]
- Total ions produced: \(2 + 1 = 3\)
- \(i = 3\)

### Step 3: Compare the values of \(i\)
Now we compare the van 't Hoff factors:
- For Al₂(SO₄)₃, \(i = 5\)
- For BaCl₂, \(i = 3\)
- For Na₂SO₄, \(i = 3\)

### Step 4: Determine the highest osmotic pressure
Since osmotic pressure is directly proportional to the van 't Hoff factor \(i\), the salt with the highest \(i\) will have the highest osmotic pressure.

- Al₂(SO₄)₃ has the highest \(i\) value of 5, compared to BaCl₂ and Na₂SO₄ which both have \(i = 3\).

### Conclusion
Thus, the salt that will have the highest osmotic pressure is **Al₂(SO₄)₃**.

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