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A 0.2 m aqueous solution of KCl freezes at -0.68°C. Calculate ‘i’ and the osmotic pressure at 0°C. Assume the volume of solution to be that of pure H2O and Kf for H2O is 1.86 K Kg/mol. |
Answer» (ΔTf)normal = Kf x m = 1.86 x 0.2= 0.372 i= Observed colligative property/Normal colligative property = 0.68/0.372 Observed osmotic pressure= i x Normal osmotic pressure = i x c RT = 1.83 x 0.2 x 0.082 x 273 = 8.2 atm |
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