1.24 g of $ {AX}_2 $ (molar mass 124 g mol$^{-1}$) is dissolved in 1 kg of water to form a solution with boiling point of 100.105$^\circ$C, while 2.54 g of $ {AY}_2 $ (molar mass 250 g mol$^{-1}$) in 2 kg of water constitutes a solution with a boiling point of 100.026$^\circ$C. $ K_{b(H_2O)} = 0.52 \, \text{K kg mol}^{-1} $. Which of the following is correct?
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The van’t Hoff factor \( i \) tells us how many particles a compound dissociates into in solution. For fully ionised compounds, \( i \) corresponds to the number of ions formed in solution.
\( {AX}_2 \) and \( {AY}_2 \) (both) are fully ionised.
\( {AX}_2 \) is fully ionised while \( {AY}_2 \) is completely unionised.
\( {AX}_2 \) and \( {AY}_2 \) (both) are completely unionised.
\( {AX}_2 \) is completely unionised while \( {AY}_2 \) is fully ionised.
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The Correct Option isA
Solution and Explanation
The elevation in boiling point is determined using the formula: \[\Delta T_b = K_b \times m \times i\] where \( \Delta T_b \) represents the boiling point elevation, \( K_b \) is the ebullioscopic constant, \( m \) is the molality, and \( i \) is the van't Hoff factor, indicating the number of particles a compound dissociates into. Given the observed changes in boiling point, we can compare the values of \( i \) (the ionization factor) for each compound.The change in boiling point is calculated as: \[\Delta T_b = T_{{solution}} - T_{{solvent}} = {100.105}^\circ C - {100.000}^\circ C = 0.105^\circ C\]For AX\(_2\):\[m = \frac{1.24 \, {g}}{124 \, {g/mol} \times 1 \, {kg H}_2{O}} = \frac{1.24}{124} \approx 0.01 \, {mol/kg}\]Applying the formula for \( \Delta T_b \):\[0.105 = 0.52 \times 0.01 \times i\]Solving for \( i \):\[i = \frac{0.105}{0.52 \times 0.01} \approx 2\]Therefore, AX\(_2\) is fully ionized (\( i = 2 \)).For AY\(_2\):\[m = \frac{2.54 \, {g}}{250 \, {g/mol} \times 2 \, {kg H}_2{O}} = \frac{2.54}{250 \times 2} \approx 0.005 \, {mol/kg}\]Using the same formula for \( \Delta T_b \):\[0.026 = 0.52 \times 0.005 \times i\]Solving for \( i \):\[i = \frac{0.026}{0.52 \times 0.005} \approx 1\]Thus, AY\(_2\) is also fully ionized (\( i = 1 \)).Consequently, both AX\(_2\) and AY\(_2\) are fully ionized.
