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Step 1: Write the Ionization Equation
The electrolyte A_2B_3 ionizes according to the following equation:
A_2B_3 \rightarrow 2A^{3+} + 3B^{2-}
Step 2: Determine the Number of Ions Produced
From the above equation, 1 formula unit of A_2B_3 produces 5 ions (2 of A^{3+} and 3 of B^{2-} ).
Step 3: Calculate the Van't Hoff Factor (i)
The van't Hoff factor i can be found using the expression:
i = 1 + (n - 1)\alpha,
where n is the total number of particles formed on complete ionization (here, n = 5 ), and \alpha is the degree of ionization (given as 0.6 ). Substituting these values:
i = 1 + (5 - 1) \times 0.6 \\
i = 1 + 4 \times 0.6 \\
i = 1 + 2.4 \\
i = 3.4
Step 4: Calculate the Elevation in Boiling Point (\Delta T_b)
Elevation in boiling point is given by:
\Delta T_b = K_b \times m \times i,
where K_b is the ebullioscopic constant of water ( 0.52\,\text{K kg mol}^{-1} ), m is the molality of the solution (given as 1\,\text{mol kg}^{-1} ), and i is the van't Hoff factor. Substituting the values:
\Delta T_b = 0.52 \times 1 \times 3.4 = 1.768\,^\circ \text{C}
Step 5: Determine the Boiling Point of the Solution
The normal boiling point of water is 100\,^\circ \text{C} (or 373\,\text{K} ). Hence, the boiling point of this solution is:
T_{\text{solution}}
= T_{\text{water}} + \Delta T_b
= 100 + 1.768
= 101.768\,^\circ \text{C}
\approx 375\,\text{K}
Thus, the boiling point of the solution at 1\,\text{atm} is approximately 375\,\text{K}.
