The freezing point of 4% aqueous solution of 'A' is equal to the freezing point 10% aqueous solution of 'B'. If the molecular mass of 'A' is 60, then the molecular mass of 'B' will be:

Options

A. 160
B. 90
C. 45
D. 180

Answer

To find the molecular mass of substance B given the freezing point depression of two solutions, we can follow these steps:

### Step 1: Understand the relationship between freezing point depression and molality
The freezing point depression (ΔTf) is given by the formula:
\[ \Delta T_f = K_f \cdot i \cdot m \]
Where:
- \( K_f \) is the freezing point depression constant,
- \( i \) is the van 't Hoff factor (which we will ignore since the question does not mention electrolytes),
- \( m \) is the molality of the solution.

Since the freezing points of the two solutions are equal, we can set their ΔTf values equal to each other:
\[ \Delta T_f(A) = \Delta T_f(B) \]

### Step 2: Set up the molality expressions for both solutions
For solution A (4% aqueous):
- Weight of solute (A) = 4 g
- Weight of solvent = 100 g - 4 g = 96 g

The molality (m) of A is given by:
\[ m_A = \frac{W_A}{M_A \cdot W_{solvent}} = \frac{4 \, \text{g}}{60 \, \text{g/mol} \cdot 0.096 \, \text{kg}} \]

For solution B (10% aqueous):
- Weight of solute (B) = 10 g
- Weight of solvent = 100 g - 10 g = 90 g

The molality (m) of B is given by:
\[ m_B = \frac{W_B}{M_B \cdot W_{solvent}} = \frac{10 \, \text{g}}{M_B \cdot 0.090 \, \text{kg}} \]

### Step 3: Set the molalities equal to each other
Since \( m_A = m_B \):
\[ \frac{4}{60 \cdot 0.096} = \frac{10}{M_B \cdot 0.090} \]

### Step 4: Cross-multiply and solve for \( M_B \)
Cross-multiplying gives:
\[ 4 \cdot M_B \cdot 0.090 = 10 \cdot 60 \cdot 0.096 \]

Now, simplify:
\[ 0.36 \cdot M_B = 576 \]

Now, solve for \( M_B \):
\[ M_B = \frac{576}{0.36} \]
\[ M_B = 1600 \, \text{g/mol} \]

### Step 5: Final answer
The molecular mass of substance B is:
\[ M_B = 160 \, \text{g/mol} \]