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.