`K_4[Fe(CN)_6]` is supposed to be 40% dissociated when 1M solution prepared. Its boiling point is equal to another 20% mass by volume of non-electrolytic solution A. Considering motality= molarity. The molecular weight of A is

Options

A. 77
B. 67
C. 57
D. 47

Answer

To solve the problem, we need to follow these steps:

### Step 1: Determine the dissociation factor (i) for K₄[Fe(CN)₆]
Given that K₄[Fe(CN)₆] is 40% dissociated in a 1M solution, we can calculate the dissociation factor (i).

1. The formula for the dissociation factor is:
\[ i = 1 + (n - 1) \cdot \alpha \]
where:
- \( n \) is the number of particles the solute dissociates into,
- \( \alpha \) is the degree of dissociation (0.4 in this case).

2. For K₄[Fe(CN)₆], it dissociates into 5 ions:
- 4 K⁺ ions and 1 [Fe(CN)₆]⁴⁻ ion, so \( n = 5 \).

3. Plugging in the values:
\[ i = 1 + (5 - 1) \cdot 0.4 = 1 + 4 \cdot 0.4 = 1 + 1.6 = 2.6 \]

### Step 2: Calculate the molarity of solution A
The problem states that a 20% mass by volume solution of A has the same boiling point elevation as the K₄[Fe(CN)₆] solution.

1. A 20% mass by volume solution means:
- 20 g of A is present in 100 mL of solution.
- Therefore, in 1000 mL (1 L), there will be:
\[ 200 \text{ g of A} \]

2. The molarity (M) of solution A can be expressed as:
\[ M = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \cdot \text{volume (L)}} \]
Thus:
\[ M = \frac{200 \text{ g}}{M_w \cdot 1 \text{ L}} = \frac{200}{M_w} \]

### Step 3: Relate boiling point elevation of both solutions
The boiling point elevation (\( \Delta T_b \)) for both solutions can be equated:

1. For K₄[Fe(CN)₆]:
\[ \Delta T_b = i \cdot m \cdot K_b \]
where \( m = 1 \) (since molality = molarity) and \( K_b \) is the ebullioscopic constant.

2. For solution A:
\[ \Delta T_b = i_A \cdot m_A \cdot K_b \]

Since both solutions have the same boiling point elevation:
\[ 2.6 \cdot 1 \cdot K_b = i_A \cdot \left(\frac{200}{M_w}\right) \cdot K_b \]

### Step 4: Solve for the molecular weight (M_w) of A
1. Cancel \( K_b \) from both sides:
\[ 2.6 = i_A \cdot \frac{200}{M_w} \]

2. Rearranging gives:
\[ M_w = \frac{200 \cdot i_A}{2.6} \]

3. Since A is a non-electrolyte, \( i_A = 1 \):
\[ M_w = \frac{200 \cdot 1}{2.6} = \frac{200}{2.6} \approx 76.92 \text{ g/mol} \]

### Final Answer
The molecular weight of A is approximately **77 g/mol**.

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