An aqueous solution of a non electrolyte is having an osmotic pressure of 2 atmosphere at 300 K. Calculate its freezing point. (Given Kf = 1.86 Kg mol-1, R = 0.821 litre-atm K-1 mol-1)
Answer & explanation
Correct answer: option 2
For a non-electrolyte
\[\pi = MRT\]
∵π=2 atm, R=0.0821 L atm K−1mol−1, T=300K
\(\text{Therefore, }M = \frac{\pi}{RT}\)
\(M = \frac{2}{0.0821 × 300} = 0.0812\)
For a very dilute solution, Molality and molarity are same
\(\Delta T_f = K_f m\)
\(\Delta T_f = 1.86 × 0.0812 =0.151\)
\(\text{Freezing point} = T_o − \Delta T\)
\(0 - 0.151\)
\(= 0.151^oC\)