What will be the osmotic pressure of 0.03N solution of Aluminium sulphate solution at `27^(@)`C ? If in solution salt dissociates 90%

Options

A. 0.566 atm
B. 0.677 atm
C. 3.399 atm
D. 4.069 atm

Explanation

To find the osmotic pressure of a 0.03N solution of Aluminium sulfate at 27°C, we can follow these steps:

### Step 1: Understand the formula for osmotic pressure
The osmotic pressure (π) is given by the formula:
\[ \pi = iCRT \]
Where:
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( C \) = concentration of the solution (in molarity)
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

### Step 2: Convert the temperature to Kelvin
The temperature given is 27°C. To convert this to Kelvin:
\[ T(K) = 27 + 273 = 300 \, K \]

### Step 3: Determine the concentration in molarity
The normality (N) of Aluminium sulfate (Al₂(SO₄)₃) is given as 0.03N. For Aluminium sulfate, which dissociates into 5 ions (2 Al³⁺ and 3 SO₄²⁻), the molarity (C) can be calculated as:
\[ C = \frac{0.03 \, \text{N}}{n} \]
Where \( n \) is the number of equivalents. Since Aluminium sulfate provides 5 ions, we have:
\[ C = 0.03 \, \text{mol/L} \]

### Step 4: Calculate the van 't Hoff factor (i)
Aluminium sulfate dissociates as follows:
\[ \text{Al}_2(\text{SO}_4)_3 \rightarrow 2 \text{Al}^{3+} + 3 \text{SO}_4^{2-} \]
Thus, it dissociates into a total of 5 ions. Given that it dissociates 90%, we can express this as:
\[ \alpha = 0.9 \]
Using the formula for the van 't Hoff factor:
\[ i = 1 + \alpha(n - 1) \]
Where \( n = 5 \):
\[ i = 1 + 0.9(5 - 1) = 1 + 0.9 \times 4 = 1 + 3.6 = 4.6 \]

### Step 5: Substitute values into the osmotic pressure formula
Now we can substitute the values into the osmotic pressure formula:
\[ \pi = iCRT \]
\[ \pi = 4.6 \times 0.03 \, \text{mol/L} \times 0.0821 \, \text{L·atm/(K·mol)} \times 300 \, K \]

### Step 6: Calculate the osmotic pressure
Calculating the above expression:
\[ \pi = 4.6 \times 0.03 \times 0.0821 \times 300 \]
\[ \pi = 4.6 \times 0.03 \times 24.63 \]
\[ \pi \approx 3.39 \, \text{atm} \]

### Final Answer
The osmotic pressure of the solution is approximately **3.39 atm**.

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