Osomotic pressure at `300 K` when `1g` glucose `(P_(1)),1g` urea`(P_(2))` and `1g` sucrose `(P_(3))` are dissolved in `500 ml` of water are follows the order.

Options

A. `P_(1)gt P_(2)gtP_(3)`
B. `P_(2)gt P_(1)gtP_(3)`
C. `P_(3)gt P_(1)gtP_(2)`
D. `P_(2)gt P_(3)gtP_(1)`

Explanation

To solve the problem regarding the osmotic pressure of solutions of glucose, urea, and sucrose, we will follow these steps:

### Step 1: Understand the Formula for Osmotic Pressure
The osmotic pressure (π) can be calculated using the formula:
\[ \pi = M \cdot R \cdot T \]
where:
- \( M \) = molarity of the solution
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Recognize Constants
In this problem, we are given that:
- The temperature \( T = 300 \, K \)
- The volume of the solution is \( 500 \, ml \) (which is \( 0.5 \, L \))
- The gas constant \( R \) is also constant.

Since \( R \) and \( T \) are constants, we can conclude that the osmotic pressure is directly proportional to the molarity \( M \).

### Step 3: Define Molarity
Molarity \( M \) is defined as:
\[ M = \frac{\text{number of moles of solute}}{\text{volume of solution in liters}} \]
The number of moles of solute can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]

### Step 4: Calculate Molar Masses
Now, we need to find the molar masses of the solutes:
- **Glucose (C6H12O6)**: Molar mass = 180 g/mol
- **Urea (NH2CONH2)**: Molar mass = 60 g/mol
- **Sucrose (C12H22O11)**: Molar mass = 342 g/mol

### Step 5: Calculate Number of Moles for Each Solute
Given that we have 1 g of each solute:
- For glucose:
\[ \text{Number of moles of glucose} = \frac{1 \, g}{180 \, g/mol} = \frac{1}{180} \, mol \]
- For urea:
\[ \text{Number of moles of urea} = \frac{1 \, g}{60 \, g/mol} = \frac{1}{60} \, mol \]
- For sucrose:
\[ \text{Number of moles of sucrose} = \frac{1 \, g}{342 \, g/mol} = \frac{1}{342} \, mol \]

### Step 6: Calculate Molarity for Each Solution
Now we can calculate the molarity \( M \) for each solution using the volume of 0.5 L:
- For glucose:
\[ M_1 = \frac{\frac{1}{180} \, mol}{0.5 \, L} = \frac{1}{90} \, mol/L \]
- For urea:
\[ M_2 = \frac{\frac{1}{60} \, mol}{0.5 \, L} = \frac{1}{30} \, mol/L \]
- For sucrose:
\[ M_3 = \frac{\frac{1}{342} \, mol}{0.5 \, L} = \frac{1}{684} \, mol/L \]

### Step 7: Determine the Order of Osmotic Pressure
Since osmotic pressure is directly proportional to molarity, we can compare the molarities to determine the order of osmotic pressure:
- \( M_1 (glucose) = \frac{1}{90} \)
- \( M_2 (urea) = \frac{1}{30} \)
- \( M_3 (sucrose) = \frac{1}{684} \)

Thus, the order of osmotic pressure is:
\[ P_2 > P_1 > P_3 \]
This means that the osmotic pressure follows the order: **urea > glucose > sucrose**.

### Final Answer
The correct order of osmotic pressure is:
\[ P_2 > P_1 > P_3 \]