Question

A sample of water is found to contain 5.85% \((\frac{w}{w})\) of AB (molecular mass 58.5) and 9.50% \((\frac{w}{w})\) XY₂ (molecular mass 95). Assuming 80% ionisation of AB and 60% ionisation of XY₂ , the freezing point of water sample is
[Given : K_f for water 1.86 K kg mol^-1 , Freezing point of pure water is 273 K and A, B and Y are monovalent ions]

• A. 264.25 K
• B. 265.56 K
• C. 280.44 K
• D. 281.75 K

Answer 1

The Correct Option is A

Freezing Point Depression Calculation 

A sample of water is found to contain 5.85% \((\frac{w}{w})\) of \(AB\) (molecular mass 58.5) and 9.50% \((\frac{w}{w})\) \(XY_2\) (molecular mass 95). Assuming 80% ionisation of \(AB\) and 60% ionisation of \(XY_2\), the freezing point of water sample is calculated as follows:

Step 1: Calculate the Molality of Each Solute

For \(AB\) (Molecular Mass 58.5):

• Weight fraction \(w = 5.85\%\)
• Molecular mass \(M = 58.5 \, \text{g/mol}\)
• Ionization percentage = 80%, so \(i = 2\)

\[ m_{AB} = \frac{0.0585 \cdot 1000}{58.5} = 1 \, \text{mol/kg} \]

For \(XY_2\) (Molecular Mass 95):

• Weight fraction \(w = 9.50\%\)
• Molecular mass \(M = 95 \, \text{g/mol}\)
• Ionization percentage = 60%, so \(i = 3\)

\[ m_{XY_2} = \frac{0.095 \cdot 1000}{95} \approx 1 \, \text{mol/kg} \]

Step 2: Calculate the Total Freezing Point Depression

\[ \Delta T_f = (i_{AB} \cdot K_f \cdot m_{AB}) + (i_{XY_2} \cdot K_f \cdot m_{XY_2}) \] \[ \Delta T_f = (2 \cdot 1.86 \cdot 1) + (3 \cdot 1.86 \cdot 1) = 3.72 + 5.58 = 9.3 \, \text{K} \]

Step 3: Calculate the Freezing Point of the Solution

The freezing point of pure water is 273 K. Therefore, the freezing point of the solution is: \[ T_f = 273 \, \text{K} - 9.3 \, \text{K} = 263.7 \, \text{K} \]

Conclusion

The closest option to our calculated value is (A) 264.25 K.

Answer 2

To calculate the freezing point depression, we use the formula: \[ \Delta T_f = K_f \times m \times i \] where: - \( \Delta T_f \) is the freezing point depression, - \( K_f \) is the cryoscopic constant (1.86 K kg mol\(^{-1}\)), - \( m \) is the molality of the solution, - \( i \) is the van't Hoff factor (which accounts for the number of particles in solution due to dissociation).

Step 1: Calculate the molality of the solution. Molality \( m \) is given by: \[ m = \frac{\text{mol of solute}}{\text{mass of solvent in kg}} \] We are given that the sample contains: - 5.85% of AB, with a molecular mass of 58.5. - 9.50% of XY₂, with a molecular mass of 95. - The solution is in water, so we assume 100 g of solution for simplicity. For AB: - Mass of AB in 100 g solution = 5.85 g - Moles of AB = \( \frac{5.85}{58.5} = 0.1 \, \text{mol} \) For XY₂: - Mass of XY₂ in 100 g solution = 9.50 g - Moles of XY₂ = \( \frac{9.50}{95} = 0.1 \, \text{mol} \) Thus, the mass of solvent (water) is: \[ 100 - (5.85 + 9.50) = 84.65 \, \text{g} = 0.08465 \, \text{kg} \] Molality of the solution is: \[ m = \frac{0.1 \, \text{mol AB} + 0.1 \, \text{mol XY₂}}{0.08465 \, \text{kg}} = 2.36 \, \text{mol/kg} \]

Step 2: Calculate the van’t Hoff factor \( i \). - For AB, 80% ionization means it dissociates into two ions (A⁺ and B⁻), so \( i = 2 \times 0.80 = 1.6 \). - For XY₂, 60% ionization means it dissociates into three ions (X⁺ and 2Y⁻), so \( i = 3 \times 0.60 = 1.8 \). The average \( i \) for the solution is: \[ i = \frac{1.6 + 1.8}{2} = 1.7 \]

Step 3: Calculate the freezing point depression. Now, we can calculate \( \Delta T_f \): \[ \Delta T_f = 1.86 \, \text{K kg mol}^{-1} \times 2.36 \, \text{mol/kg} \times 1.7 = 7.53 \, \text{K} \]

Step 4: Calculate the freezing point of the solution. The freezing point of pure water is 273 K, so the freezing point of the solution is: \[ 273 \, \text{K} - 7.53 \, \text{K} = 264.25 \, \text{K} \]

Thus, the freezing point of the water sample is: \[{\text{(A) 264.25 K}} \]