A 1 molal $$K_4Fe(CN)_6$$ solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of A is ________ u. (Round off to the Nearest Integer). [Density of water = 1.0 g cm$$^{-3}$$]

For the $$\text{K}_4[\text{Fe}(\text{CN})_6]$$ solution, the van't Hoff factor $$i$$ needs to be calculated. This salt dissociates as $$\text{K}_4[\text{Fe}(\text{CN})_6] \to 4\text{K}^+ + [\text{Fe}(\text{CN})_6]^{4-}$$, giving a total of 5 ions per formula unit.

With a degree of dissociation $$\alpha = 0.4$$, the van't Hoff factor is $$i = 1 + (n - 1)\alpha = 1 + (5 - 1)(0.4) = 1 + 1.6 = 2.6$$.

The boiling point elevation for a 1 molal solution of the electrolyte is $$\Delta T_{b,1} = i \cdot K_b \cdot m = 2.6 \times K_b \times 1 = 2.6\, K_b$$.

For the non-electrolytic solute A, the solution contains 18.1 weight percent of A, meaning 18.1 g of A is dissolved in 81.9 g of water. The molality of this solution is:

$$m_A = \frac{18.1/M_A}{81.9/1000} = \frac{18.1 \times 1000}{M_A \times 81.9} = \frac{18100}{81.9\, M_A}$$

Since A is a non-electrolyte, $$i = 1$$, and the boiling point elevation is $$\Delta T_{b,2} = K_b \times \frac{18100}{81.9\, M_A}$$.

Setting the two boiling point elevations equal (since the boiling points are the same):

$$2.6\, K_b = K_b \times \frac{18100}{81.9\, M_A}$$

Cancelling $$K_b$$ from both sides:

$$2.6 = \frac{18100}{81.9\, M_A}$$

$$M_A = \frac{18100}{81.9 \times 2.6} = \frac{18100}{212.94} = 85.0 \text{ u}$$

The molar mass of solute A is 85 u.

The answer is 85.