1. Use boiling point elevation relation

For any solution, $\Delta T_b = i K_b m$

If two solutions have the same boiling point, then their boiling point elevations are equal: $i_1 m_1 = i_2 m_2$

For the non-electrolyte, $i_2 = 1$

2. Find van’t Hoff factor for $K_4Fe(CN)_6$

The salt dissociates as $K_4Fe(CN)_6 \rightarrow 4K^+ + [Fe(CN)_6]^{4-}$

So, total number of ions formed on complete dissociation = $5$.

Given degree of dissociation, $\alpha = 0.4$

Hence, $i = 1 + (5-1)\alpha$ $i = 1 + 4(0.4) = 2.6$

Given molality of this solution: $m_1 = 1$

Thus, $i_1 m_1 = 2.6 \times 1 = 2.6$

So for the second solution, $m_2 = 2.6$

3. Interpret 18.1 wt% solution

Take $100\,$g of solution.

Then:

• mass of solute $A = 18.1\,$g
• mass of solvent (water) $= 81.9\,$g $= 0.0819\,$kg

Let molar mass of $A$ be $M$.

Then moles of $A$ are $\frac{18.1}{M}$

Molality is $m_2 = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{18.1/M}{0.0819}$

Given this equals $2.6$: $\frac{18.1/M}{0.0819} = 2.6$

4. Solve for $M$

$\frac{18.1}{0.0819 M} = 2.6$

$18.1 = 2.6 \times 0.0819 \times M$

$M = \frac{18.1}{2.6 \times 0.0819}$

$2.6 \times 0.0819 = 0.21294$

$M = \frac{18.1}{0.21294} \approx 84.9986$

Thus, $M \approx 85\ \text{u}$

5. Final Answer

The molar mass of solute $A$ is $\boxed{85\ \text{u}}$