The `1.25` molal sucrose solution at temperature `60^(@)` has a density of `0.1142 g//ml`, then the osmotic pressure of solution will be

Options

A. `2.5`
B. `3.5`
C. `2.0`
D. `4.0`

Explanation

To find the osmotic pressure of a 1.25 molal sucrose solution at 60°C with a density of 0.1142 g/mL, we can follow these steps:

### Step 1: Convert the temperature to Kelvin
The temperature in Kelvin (K) can be calculated using the formula:
\[ T(K) = T(°C) + 273 \]
Given \( T = 60°C \):
\[ T = 60 + 273 = 333 \, K \]

### Step 2: Use the relationship between molality and molarity
The relationship between molality (m) and molarity (M) is given by:
\[ \frac{1}{m} = \frac{D}{M} - \frac{M_{solute}}{1000} \]
Where:
- \( m = 1.25 \, \text{molal} \)
- \( D = 0.1142 \, \text{g/mL} = 114.2 \, \text{g/L} \) (since 1 mL = 1 g)
- \( M_{solute} = 342 \, \text{g/mol} \) (molecular mass of sucrose)

### Step 3: Rearranging the equation to find molarity (M)
Rearranging the equation gives:
\[ \frac{1}{1.25} = \frac{114.2}{M} - \frac{342}{1000} \]
This can be rewritten as:
\[ \frac{114.2}{M} = \frac{1}{1.25} + \frac{342}{1000} \]
Calculating \( \frac{1}{1.25} \):
\[ \frac{1}{1.25} = 0.8 \]
Calculating \( \frac{342}{1000} \):
\[ \frac{342}{1000} = 0.342 \]
So we have:
\[ \frac{114.2}{M} = 0.8 + 0.342 = 1.142 \]

### Step 4: Solve for M
Now, we can solve for \( M \):
\[ M = \frac{114.2}{1.142} \approx 100.0 \, \text{mol/L} \]

### Step 5: Calculate the osmotic pressure (π)
The osmotic pressure can be calculated using the formula:
\[ \pi = I \cdot M \cdot R \cdot T \]
Where:
- \( I = 1 \) (for sucrose, a non-electrolyte)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 333 \, K \)

Substituting the values:
\[ \pi = 1 \cdot 0.1 \cdot 0.0821 \cdot 333 \]
Calculating:
\[ \pi \approx 2.73 \, \text{atm} \]

### Step 6: Round to the nearest option
The closest value to 2.73 atm is approximately 2.5 atm.

### Final Answer
The osmotic pressure of the solution is approximately **2.5 atm**.

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