Read the paragraph carefully and answer the following questions:
Mole fraction of acetic acid in benzene given a solution having freezing point 277.4K is 0.02. Freezing point of benezene is 278.4K and heat of fusion of benzene is 10.042 kJ/mol. If molarity of solution is equal to molality, and acetic acid is found in equilibrium with it dimmer than answer the following based upon above data
What should be the molality of the solution formed

Options

A. 0.462 m
B. 0.4 m
C. 0.262 m
D. 0.626 m

Answer

To find the molality of the solution formed, we can follow these steps:

### Step 1: Understand the given data
- Mole fraction of acetic acid (solute) in benzene (solvent) = 0.02
- Freezing point of benzene = 278.4 K
- Freezing point of the solution = 277.4 K
- Heat of fusion of benzene = 10.042 kJ/mol
- Molarity = Molality (given)

### Step 2: Calculate the mole fraction of the solvent (benzene)
The mole fraction of the solvent can be calculated using the formula:
\[ x_{\text{solvent}} = 1 - x_{\text{solute}} \]
Substituting the value of the mole fraction of acetic acid:
\[ x_{\text{solvent}} = 1 - 0.02 = 0.98 \]

### Step 3: Use the relationship between molality and mole fraction
The formula for molality (m) in terms of mole fraction is given by:
\[ m = \frac{x_{\text{solute}} \times 1000}{x_{\text{solvent}} \times M_{\text{solvent}}} \]
where \(M_{\text{solvent}}\) is the molar mass of the solvent (benzene).

### Step 4: Find the molar mass of benzene
The molar mass of benzene (C6H6) is approximately 78 g/mol.

### Step 5: Substitute the values into the molality formula
Substituting the known values into the molality formula:
\[ m = \frac{0.02 \times 1000}{0.98 \times 78} \]

### Step 6: Calculate the molality
Now, calculate the molality:
\[ m = \frac{20}{76.44} \approx 0.2617 \text{ mol/kg} \]

### Step 7: Round the answer
Rounding to three significant figures, we get:
\[ m \approx 0.262 \text{ mol/kg} \]

### Final Answer
The molality of the solution formed is approximately **0.262 mol/kg**.

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