The depession inn freezing point of a 5% aqueous solution of a substance 'A' is equal to the depression in freezing point of a 3% aqueous solution of a substance 'B' .If the molecules weight of'A' is ________.

Options

A. `72 g//mol`
B. `112.2 g//mol`
C. `117.52 g//mol`
D. `122.52 g//mol`

Answer

To solve the problem, we need to find the molecular weight of substance 'B' given that the depression in freezing point of a 5% aqueous solution of substance 'A' is equal to the depression in freezing point of a 3% aqueous solution of substance 'B'. We will use the formula for depression in freezing point and the concept of molality.

### Step-by-Step Solution:

1. **Understanding the Depression in Freezing Point**:
The depression in freezing point (ΔTf) can be calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \(i\) = van 't Hoff factor (number of particles the solute breaks into)
- \(K_f\) = freezing point depression constant of the solvent (water in this case)
- \(m\) = molality of the solution

2. **Setting Up the Equation**:
Since the depression in freezing point for both solutions is equal, we can set up the equation:
\[ \Delta T_f(A) = \Delta T_f(B) \]
This implies:
\[ i_A \cdot K_f \cdot m_A = i_B \cdot K_f \cdot m_B \]
Given that both solutes are non-electrolytes (assuming \(i_A = i_B = 1\)), we can simplify this to:
\[ m_A = m_B \]

3. **Calculating Molality for Substance A**:
For substance A:
- A 5% solution means 5 g of A in 100 g of solution.
- The mass of the solvent (water) = 100 g - 5 g = 95 g = 0.095 kg.
- The molality \(m_A\) is given by:
\[ m_A = \frac{\text{moles of A}}{\text{mass of solvent in kg}} = \frac{\frac{5}{M_A}}{0.095} \]
where \(M_A\) is the molecular weight of A.

4. **Calculating Molality for Substance B**:
For substance B:
- A 3% solution means 3 g of B in 100 g of solution.
- The mass of the solvent (water) = 100 g - 3 g = 97 g = 0.097 kg.
- The molality \(m_B\) is given by:
\[ m_B = \frac{\text{moles of B}}{\text{mass of solvent in kg}} = \frac{\frac{3}{M_B}}{0.097} \]
where \(M_B\) is the molecular weight of B.

5. **Setting the Molalities Equal**:
Since \(m_A = m_B\), we can set the two expressions for molality equal:
\[ \frac{\frac{5}{M_A}}{0.095} = \frac{\frac{3}{M_B}}{0.097} \]

6. **Cross Multiplying to Solve for \(M_B\)**:
Cross-multiplying gives:
\[ 5 \cdot 0.097 \cdot M_B = 3 \cdot 0.095 \cdot M_A \]
Rearranging this gives:
\[ M_B = \frac{3 \cdot 0.095 \cdot M_A}{5 \cdot 0.097} \]

7. **Substituting the Value of \(M_A\)**:
Given that \(M_A = 208.72 \, \text{g/mol}\):
\[ M_B = \frac{3 \cdot 0.095 \cdot 208.72}{5 \cdot 0.097} \]

8. **Calculating \(M_B\)**:
Performing the calculations:
\[ M_B = \frac{0.2856 \cdot 208.72}{0.485} \approx 122.52 \, \text{g/mol} \]

### Final Answer:
The molecular weight of substance B is approximately **122.52 g/mol**.