The relationship between osmotic pressure (`pi_(1),pi_(2) "and" pi_(3)`) at a definite temperature when 1 g glucose, 1 g urea and 1 g sucrose are dissolved in 1 litre of water is (assume i = 1 for all):

Options

A. (a) `pi_(1) gt pi_(2) gt pi_(3)`
B. (b) `pi_(3) gt pi_(1) gt pi_(2)`
C. (c) `pi_(2) gt pi_(1) gt pi_(3)`
D. (d) `pi_(2) gt pi_(3) gt pi_(1)`

Answer

To solve the problem, we need to determine the relationship between the osmotic pressures (\( \pi_1, \pi_2, \pi_3 \)) of glucose, urea, and sucrose when 1 gram of each is dissolved in 1 liter of water. We'll use the formula for osmotic pressure, which is given by:

\[ \pi = iCRT \]

Where:
- \( \pi \) = osmotic pressure
- \( i \) = Van't Hoff factor (which is 1 for all three substances in this case)
- \( C \) = concentration (in moles per liter)
- \( R \) = universal gas constant (constant for all)
- \( T \) = temperature (constant for all)

### Step 1: Calculate the number of moles of each solute

The number of moles (\( n \)) can be calculated using the formula:

\[ n = \frac{\text{mass}}{\text{molar mass}} \]

#### For Glucose (C₆H₁₂O₆):
- Molar mass of glucose = \( 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \, \text{g/mol} \)
- Number of moles of glucose (\( n_1 \)):
\[ n_1 = \frac{1 \, \text{g}}{180 \, \text{g/mol}} = \frac{1}{180} \, \text{mol} \]

#### For Urea (CH₄N₂O):
- Molar mass of urea = \( 12 + 4 \times 1 + 14 \times 2 + 16 = 60 \, \text{g/mol} \)
- Number of moles of urea (\( n_2 \)):
\[ n_2 = \frac{1 \, \text{g}}{60 \, \text{g/mol}} = \frac{1}{60} \, \text{mol} \]

#### For Sucrose (C₁₂H₂₂O₁₁):
- Molar mass of sucrose = \( 12 \times 12 + 22 \times 1 + 11 \times 16 = 342 \, \text{g/mol} \)
- Number of moles of sucrose (\( n_3 \)):
\[ n_3 = \frac{1 \, \text{g}}{342 \, \text{g/mol}} = \frac{1}{342} \, \text{mol} \]

### Step 2: Calculate the osmotic pressures

Using the osmotic pressure formula, we can express the osmotic pressures for each solute:

\[ \pi_1 = i \cdot C_1 \cdot R \cdot T = i \cdot \frac{n_1}{V} \cdot R \cdot T \]
\[ \pi_2 = i \cdot C_2 \cdot R \cdot T = i \cdot \frac{n_2}{V} \cdot R \cdot T \]
\[ \pi_3 = i \cdot C_3 \cdot R \cdot T = i \cdot \frac{n_3}{V} \cdot R \cdot T \]

Since \( i \), \( R \), \( T \), and \( V \) are constants and equal for all three, we can compare the osmotic pressures based on the number of moles:

\[ \frac{\pi_1}{\pi_2} = \frac{n_1}{n_2}, \quad \frac{\pi_1}{\pi_3} = \frac{n_1}{n_3}, \quad \frac{\pi_2}{\pi_3} = \frac{n_2}{n_3} \]

### Step 3: Compare the number of moles

Now we can compare the number of moles we calculated:

- \( n_1 = \frac{1}{180} \)
- \( n_2 = \frac{1}{60} \)
- \( n_3 = \frac{1}{342} \)

### Step 4: Determine the order of osmotic pressures

To find the relationship between osmotic pressures:

- Since \( n_2 > n_1 > n_3 \), we have:
\[ \pi_2 > \pi_1 > \pi_3 \]

Thus, the order of osmotic pressures is:
\[ \pi_2 > \pi_1 > \pi_3 \]

### Conclusion

The correct relationship between the osmotic pressures is:
\[ \pi_2 > \pi_1 > \pi_3 \]