Question

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 K kg mol\(^{-1}\)} Which of the following is correct?

• A. \( {AX}_2 \) and \( {AY}_2 \) (both) are fully ionised.
• B. \( {AX}_2 \) is fully ionised while \( {AY}_2 \) is completely unionised.
• C. \( {AX}_2 \) and \( {AY}_2 \) (both) are completely unionised.
• D. \( {AX}_2 \) is completely unionised while \( {AY}_2 \) is fully ionised.

Hint:

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.

Answer 1

The Correct Option is A

To find the correct answer, we use the formula for the elevation in boiling point, which is given by: \[ \Delta T_b = K_b \times m \times i \] where \( \Delta T_b \) is the boiling point elevation, \( K_b \) is the ebullioscopic constant, \( m \) is the molality, and \( i \) is the van't Hoff factor (which gives the number of particles the compound dissociates into). Since we are given the boiling point changes, we can compare the values of \( i \) (the ionisation factor) for each compound. The change in boiling point is given as: \[ \Delta T_b = T_{{solution}} - T_{{solvent}} = {100.105}^\circ C - {100.000}^\circ C = 0.105^\circ C \] For AX\(_2\): \[ m = \frac{1.24 \, {g}}{124 \, {g/mol} \times 1 \, {kg H}_2{O}} = \frac{1.24}{124} \approx 0.01 \, {mol/kg} \] Using the formula for \( \Delta T_b \): \[ 0.105 = 0.52 \times 0.01 \times i \] Solving for \( i \), we get: \[ i = \frac{0.105}{0.52 \times 0.01} \approx 2 \] Thus, AX\(_2\) is fully ionised (since \( i = 2 \)). 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 \( i \), we get: \[ i = \frac{0.026}{0.52 \times 0.005} \approx 1 \] Thus, AY\(_2\) is also fully ionised (since \( i = 1 \)). Hence, the correct answer is that both AX\(_2\) and AY\(_2\) are fully ionised.