The vapour pressure of pure benzene at `25^@C` is 640 mm Hg and that of the solute A in benzene is 630 mm of Hg. The molality of solution of

Options

A. 0.2 m
B. 0.4 m
C. 0.5 m
D. 0.1 m

Answer

To find the molality of the solution, we can use the concept of relative lowering of vapor pressure. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Vapor pressure of pure benzene (P₀) = 640 mm Hg
- Vapor pressure of the solution (P) = 630 mm Hg

### Step 2: Calculate the relative lowering of vapor pressure
The formula for relative lowering of vapor pressure is given by:

\[ \Delta P = \frac{P_0 - P}{P_0} \]

Substituting the values:

\[ \Delta P = \frac{640 \, \text{mm Hg} - 630 \, \text{mm Hg}}{640 \, \text{mm Hg}} = \frac{10}{640} = 0.015625 \]

### Step 3: Relate relative lowering to the number of moles
According to Raoult's law, the relative lowering of vapor pressure is also equal to the ratio of the number of moles of solute (n₁) to the number of moles of solvent (n₂):

\[ \Delta P = \frac{n_1}{n_2} \]

### Step 4: Express moles in terms of mass
The number of moles can be expressed as:

\[ n_1 = \frac{m_1}{M_1} \quad \text{and} \quad n_2 = \frac{m_2}{M_2} \]

Where:
- \(m_1\) = mass of solute (in grams)
- \(M_1\) = molar mass of solute (in g/mol)
- \(m_2\) = mass of solvent (in grams)
- \(M_2\) = molar mass of solvent (in g/mol)

For benzene (C₆H₆), the molar mass \(M_2\) is:

\[ M_2 = 12 \times 6 + 1 \times 6 = 72 + 6 = 78 \, \text{g/mol} \]

### Step 5: Substitute into the equation
Now substituting these into the equation for relative lowering:

\[ 0.015625 = \frac{\frac{m_1}{M_1}}{\frac{m_2}{M_2}} \]

Rearranging gives:

\[ 0.015625 = \frac{m_1 \cdot M_2}{m_2 \cdot M_1} \]

### Step 6: Express molality
Molality (m) is defined as:

\[ m = \frac{m_1}{M_1 \cdot \frac{m_2}{1000}} \quad \text{(where } m_2 \text{ is in kg)} \]

Thus, we can express \(m_2\) in kg as \(m_2 = 1000 \, \text{g}\) (assuming we have 1 kg of solvent for simplicity).

### Step 7: Rearranging to find molality
Substituting \(m_2 = 1000 \, \text{g}\) into the equation:

\[ 0.015625 = \frac{m_1 \cdot 78}{1000 \cdot M_1} \]

### Step 8: Solve for molality
Now, rearranging gives:

\[ m_1 = 0.015625 \cdot \frac{1000 \cdot M_1}{78} \]

To find molality (m):

\[ m = \frac{m_1}{M_1 \cdot 1} = 0.015625 \cdot \frac{1000}{78} \]

Calculating this gives:

\[ m = 0.2 \, \text{mol/kg} \]

### Final Answer
Thus, the molality of the solution is **0.2 molal**.

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