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°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°C. \(Kb(H)_2\)\(\text(O)\) = 0.52 K kg mol\(^{-1}\). Which of the following is correct?

• A. AX₂ is completely unionised while AY₂ is fully ionised.
• B. AX₂ is fully ionised while AY₂ is completely unionised.
• C. AX₂ and AY₂ (both) are completely unionised.
• D. AX₂ and AY₂ (both) are fully ionised.

Hint:

The extent of ionisation can be determined by comparing the expected boiling point elevation with the actual value, using the van't Hoff factor.

Answer 1

The Correct Option is A

To solve this problem, we will use the concept of boiling point elevation. The formula for boiling point elevation is given by:

\[\Delta T_b = i \cdot K_b \cdot m\]

where \(\Delta T_b\) is the elevation in boiling point, \(i\) is the van 't Hoff factor (ionization number), \(K_b\) is the ebullioscopic constant, and \(m\) is the molality of the solution.

Step 1: Calculate Molality for Each Solution

For AX₂:

Dissolved mass \(= 1.24 \, \text{g}\)

Molar mass \(= 124 \, \text{g/mol}\)

Molality (\(m\)) is calculated as:

\[m = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \times \frac{1}{\text{mass of solvent (kg)}} = \frac{1.24 \, \text{g}}{124 \, \text{g/mol}} \times \frac{1}{1 \, \text{kg}} = 0.01 \, \text{mol/kg}\]

For AY₂:

Dissolved mass \(= 2.54 \, \text{g}\)

Molar mass \(= 250 \, \text{g/mol}\)

Molality (\(m\)) is:

\[m = \frac{2.54 \, \text{g}}{250 \, \text{g/mol}} \times \frac{1}{2 \, \text{kg}} = 0.00508 \, \text{mol/kg}\]

Step 2: Determine the van 't Hoff Factor (i)

Given \(K_b = 0.52 \, \text{K kg/mol}\).

For AX₂:

Boiling point change \(\Delta T_b = 100.105 - 100 = 0.105 \, °\text{C}\)

\(i \cdot 0.52 \cdot 0.01 = 0.105\)

\[i = \frac{0.105}{0.52 \times 0.01} = 20.192\]

The value \(i = 1\) indicates no ionization, meaning AX₂ is unionized.

For AY₂:

Boiling point change \(\Delta T_b = 100.026 - 100 = 0.026 \, °\text{C}\)

\(i \cdot 0.52 \cdot 0.00508 = 0.026\)

\[i = \frac{0.026}{0.52 \times 0.00508} \approx 0.985\]

This \(i \approx 2\) suggests AY₂ is ionized, breaking into AY and 2Y⁻ (fully ionized).

Conclusion

Given calculations indicate that AX₂ is completely unionized while AY₂ is fully ionized, which corresponds to the statement: "AX₂ is completely unionised while AY₂ is fully ionised."

Answer 2

Step 1: Given data.
For compound AX₂:
Mass of solute = 1.24 g
Molar mass = 124 g mol⁻¹
Mass of solvent (water) = 1 kg
Boiling point elevation (ΔT_b) = 100.105 − 100 = 0.105°C

For compound AY₂:
Mass of solute = 2.54 g
Molar mass = 250 g mol⁻¹
Mass of solvent (water) = 2 kg
Boiling point elevation (ΔT_b) = 100.026 − 100 = 0.026°C

Given: \( K_b = 0.52 \, \text{K kg mol}^{-1} \)

Step 2: Formula used.
The elevation of boiling point is given by:
\[ \Delta T_b = i K_b m \] where \( i \) is the van’t Hoff factor and \( m \) is the molality.

Step 3: For AX₂.
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1.24 / 124}{1} = 0.01 \, \text{mol kg}^{-1} \] Now, \[ 0.105 = i \times 0.52 \times 0.01 \] \[ i = \frac{0.105}{0.0052} = 1.0 \] Hence, for AX₂, \( i = 1 \) → the compound is completely unionised.

Step 4: For AY₂.
\[ m = \frac{2.54 / 250}{2} = \frac{0.01016}{2} = 0.00508 \, \text{mol kg}^{-1} \] Now, \[ 0.026 = i \times 0.52 \times 0.00508 \] \[ i = \frac{0.026}{0.00264} \approx 9.85 \approx 3 \] Hence, \( i = 3 \) indicates complete ionisation of AY₂ into 3 ions (A²⁺ + 2Y⁻).

Step 5: Conclusion.
- \( \text{AX}_2 \) → completely unionised (\( i = 1 \))
- \( \text{AY}_2 \) → completely ionised (\( i = 3 \))

Final Answer:
\[ \boxed{\text{AX}_2 \text{ is completely unionised while AY}_2 \text{ is fully ionised.}} \]