Question

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 $\frac{A}{9}$ is...............u. (Round off to the Nearest Integer).

[Density of water = 1.0 g $cm^{-3}$]

Answer 1

9

Answer 2

The boiling point elevation of a solution is given by $\Delta T_b = i \cdot K_b \cdot m$, where $i$ is the van't Hoff factor, $K_b$ is the ebullioscopic constant of the solvent, and $m$ is the molality of the solution.

For the 1 molal $K_4Fe(CN)_6$ solution: $K_4Fe(CN)_6$ dissociates in water as follows: $K_4Fe(CN)_6 \rightleftharpoons 4K^+ + [Fe(CN)_6]^{4-}$ One molecule of $K_4Fe(CN)_6$ produces $n = 4 + 1 = 5$ ions. The degree of dissociation is given as $\alpha = 0.4$. The van't Hoff factor $i$ is related to the degree of dissociation by the formula: $i = 1 + \alpha(n-1)$ $i = 1 + 0.4(5-1) = 1 + 0.4(4) = 1 + 1.6 = 2.6$. The molality of the solution is $m = 1$ molal. The boiling point elevation for the $K_4Fe(CN)_6$ solution is: $\Delta T_b(K_4Fe(CN)_6) = i \cdot K_b \cdot m = 2.6 \cdot K_b \cdot 1 = 2.6 K_b$.

For the second solution containing a non-electrolytic solute A: The solution contains 18.1 weight percent of solute A. This means that in 100 g of the solution, there are 18.1 g of solute A and $100 - 18.1 = 81.9$ g of solvent (water). To find the molality ($m_A$), we need the number of moles of A and the mass of the solvent in kilograms. Let $M_A$ be the molar mass of A in g/mol. Number of moles of A = $\frac{\text{Mass of A}}{\text{Molar mass of A}} = \frac{18.1}{M_A}$ moles. Mass of solvent (water) = 81.9 g = 0.0819 kg. Molality of solution A is: $m_A = \frac{\text{Moles of A}}{\text{Mass of solvent (kg)}} = \frac{18.1/M_A}{0.0819} = \frac{18.1}{M_A \cdot 0.0819}$. Since A is a non-electrolytic solute, its van't Hoff factor is $i_A = 1$. The boiling point elevation for solution A is: $\Delta T_b(A) = i_A \cdot K_b \cdot m_A = 1 \cdot K_b \cdot \frac{18.1}{M_A \cdot 0.0819} = K_b \cdot \frac{18.1}{M_A \cdot 0.0819}$.

The problem states that the boiling point of the $K_4Fe(CN)_6$ solution is equal to that of the solution containing A. This implies that their boiling point elevations are equal: $\Delta T_b(K_4Fe(CN)_6) = \Delta T_b(A)$ $2.6 K_b = K_b \cdot \frac{18.1}{M_A \cdot 0.0819}$ Assuming $K_b \neq 0$ (which is true for water), we can cancel $K_b$ from both sides: $2.6 = \frac{18.1}{M_A \cdot 0.0819}$ Now, we solve for $M_A$: $M_A \cdot 0.0819 = \frac{18.1}{2.6}$ $M_A = \frac{18.1}{2.6 \cdot 0.0819}$ Calculate the denominator: $2.6 \times 0.0819 = 0.21294$. $M_A = \frac{18.1}{0.21294}$ $M_A \approx 85.00047$ g/mol.

The question asks for the molar mass of $\frac{A}{9}$ in u (unified atomic mass units). The numerical value of the molar mass in g/mol is equal to the numerical value of the mass of one molecule in u. We need to calculate $\frac{M_A}{9}$: $\frac{M_A}{9} = \frac{85.00047}{9} \approx 9.444496...$

We need to round off the result to the Nearest Integer. The nearest integer to 9.444... is 9.

The final answer is 9.