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Arrange the following solutions in the increasing order of their osmotic pressure a) 34.2 g/lit sucrose (b) 60g/lit urea (NH_(2)CONH_(2)) (c) 90 g/lit glucose (d) 58.5 g/lit sodium chloride Give reason in support of your answer |
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Answer» Solution :MOLAR mass of sucrose `(C_(12)H_(22)O_(11))="342 g mol"^(-1)` Molar mass of urea `(NH_(2)CONH_(2))="60 g mol"^(-1)," Molar mass of glucose "(C_(6)H_(12)O_(6))="180 g mol"^(-1)` Molar mass of `NaCl="58.8 g mol"^(-1),` Molar conc. Of sucrose `=(34.2)/(342)=0.1M,"Molar conc of urea "=(60)/(60)=1M,` Molar conc of glucose `=(90)/(180)=0.5M,"Molar conc. of NaCl"=(58.5)/(58.5)=1M,` However, as NaCl is an electrolyte and one formula unit of NaCl dissociates to give two ions (`Na^(+) and Cl^(-)`) therefore, molar concentration of particles in the solution = 2M. Thus, the ORDER or increasing concentration is `UNDERSET("(0.1 M)")("Sucrose")""lt""underset("(0.5 M)")("Glucose")""lt""underset("(1 M)")("Urea")""lt""underset("(2 M)")("NaCl")` As osmoticpressure (or any colligative property) is directly proportional to the number of particles in the solution, hence increasing order of osmotic pressure will be : `"Sucrose" lt"Glucose"lt"Urea"lt "NaCl"` Note. If the above solutions are to be arranged in order of their decreasing freezing points, the depression in freezing points `(DeltaT_(F))` will be in the order : `"Sucrose "lt "Glucose"lt"Urea"lt"NaCl"` Thus, sucrose will have MINIMUM depression i.e., its actual freezing point will be maximum. Hence, the order of decreasing freezing points will be : `"Sucrose"gt "Glucose"gt "Urea"gt"NaCl"` |
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