The freezing point of a `3%` aqueous solution 'A' is equal to the freezing point of `9%` aqueous solution 'B'. If the molecular weight of 'A' is 60, then the molecular weight of 'B' will be

Options

A. 191.9
B. 90
C. 45
D. 20

Answer

To solve the problem, we will use the concept of freezing point depression and the relationship between molality, molecular weight, and the mass of solute.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- We have two solutions: A (3% solution) and B (9% solution).
- The molecular weight of A is given as 60 g/mol.
- The freezing points of both solutions are equal.

2. **Set Up the Freezing Point Depression Formula**:
The freezing point depression (ΔTf) can be expressed as:
\[ \Delta T_f = K_f \cdot i \cdot m \]
where:
- \( K_f \) is the cryoscopic constant (which is the same for both solutions).
- \( i \) is the van 't Hoff factor (which is 1 for non-electrolytes).
- \( m \) is the molality of the solution.

3. **Calculate the Molality of Solution A**:
- A 3% solution means there are 3 grams of solute (A) in 100 grams of solution.
- The mass of the solvent (water) is:
\[ 100 \, \text{g} - 3 \, \text{g} = 97 \, \text{g} \]
- Convert the mass of the solvent to kilograms:
\[ 97 \, \text{g} = 0.097 \, \text{kg} \]
- The molality (m) of solution A is calculated as:
\[ m_A = \frac{\text{mass of solute (A)}}{\text{molecular weight of A} \times \text{mass of solvent in kg}} = \frac{3 \, \text{g}}{60 \, \text{g/mol} \times 0.097 \, \text{kg}} = \frac{3}{5.82} \approx 0.515 \, \text{mol/kg} \]

4. **Calculate the Molality of Solution B**:
- A 9% solution means there are 9 grams of solute (B) in 100 grams of solution.
- The mass of the solvent (water) is:
\[ 100 \, \text{g} - 9 \, \text{g} = 91 \, \text{g} \]
- Convert the mass of the solvent to kilograms:
\[ 91 \, \text{g} = 0.091 \, \text{kg} \]
- The molality (m) of solution B is:
\[ m_B = \frac{9 \, \text{g}}{M_B \times 0.091 \, \text{kg}} \quad \text{(where \( M_B \) is the molecular weight of B)} \]

5. **Set the Molalities Equal**:
Since the freezing points are equal, the molalities must also be equal:
\[ m_A = m_B \]
\[ 0.515 = \frac{9}{M_B \times 0.091} \]

6. **Solve for Molecular Weight of B**:
Rearranging the equation gives:
\[ M_B = \frac{9}{0.515 \times 0.091} \]
\[ M_B = \frac{9}{0.046765} \approx 192.3 \, \text{g/mol} \]

### Final Answer:
The molecular weight of B is approximately **192.3 g/mol**.