Solute `A` is a ternary electrolyte and solute `B` is non-elctrolyte . If`0.1`M solution of solute`B` produces an osmotic pressure of `2P`,then `0.05M`solution of `A`at the same temperature will produce on osmotic pressure equal to

Options

A. `2P`
B. `1.5P`
C. `3P`
D. none of these

Explanation

To solve the problem, we will use the formula for osmotic pressure, which is given by:

\[ \pi = i \cdot m \cdot R \cdot T \]

Where:
- \(\pi\) = osmotic pressure
- \(i\) = van't Hoff factor
- \(m\) = molarity of the solution
- \(R\) = gas constant
- \(T\) = temperature

### Step 1: Calculate the osmotic pressure for solute B

Given that:
- The molarity of solute B (\(m_B\)) = 0.1 M
- The osmotic pressure of solute B (\(\pi_B\)) = 2P
- Since B is a non-electrolyte, its van't Hoff factor (\(i_B\)) = 1.

Using the osmotic pressure formula for solute B:

\[ \pi_B = i_B \cdot m_B \cdot R \cdot T \]

Substituting the known values:

\[ 2P = 1 \cdot 0.1 \cdot R \cdot T \]

This simplifies to:

\[ 2P = 0.1 \cdot R \cdot T \]

### Step 2: Express \(R \cdot T\) in terms of P

From the equation above, we can express \(R \cdot T\):

\[ R \cdot T = \frac{2P}{0.1} = 20P \]

### Step 3: Calculate the osmotic pressure for solute A

Now, we need to calculate the osmotic pressure for solute A. Given:
- The molarity of solute A (\(m_A\)) = 0.05 M
- Since A is a ternary electrolyte, its van't Hoff factor (\(i_A\)) = 3.

Using the osmotic pressure formula for solute A:

\[ \pi_A = i_A \cdot m_A \cdot R \cdot T \]

Substituting the known values:

\[ \pi_A = 3 \cdot 0.05 \cdot R \cdot T \]

### Step 4: Substitute \(R \cdot T\) into the equation for \(\pi_A\)

Now we substitute \(R \cdot T\) from Step 2 into the equation for \(\pi_A\):

\[ \pi_A = 3 \cdot 0.05 \cdot (20P) \]

Calculating this gives:

\[ \pi_A = 3 \cdot 0.05 \cdot 20P = 3 \cdot 1P = 3P \]

### Final Answer

Thus, the osmotic pressure produced by a 0.05 M solution of solute A is:

\[ \pi_A = 3P \]