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Derive the equation, alpha=(i-1)/(n-1) Derive a relation between degree of dissociation and van't Hoff factor. |
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Answer» Solution :Consider one mole of a substance An (like an electrolyte), which on dissociation in the solution gives n number of particles A. Let `alpha` be the degree of dissociation at equilibrium. Then, the dissociation equilibrium can be represented as, `An Leftrightarrow nA` `{:(,"At start",1,0,"mole"),(,"At equilibrium",1-alpha,nalpha,"mole"):} ` At start, let the number of moles of An (or particles) be 1. At equilibrium The total number of moles of particles `=1-alpha+nalpha` `=1+nalpha-alpha` `=1+alpha(n-1)` THEORETICAL COLLIGATIVE properties are DUE to 1 mole while the observed colligative properties are due to `1+alpha(n-1)` moles of particles in the solution. The van't HOFF factor i, will be, `i=("Observed colligative property")/("Theoretical colligative property")` Now, Colligative property `alpha` Number of particels in the solution. ` therefore i=(1+alpha(n-1))/(1)` `therefore hati=1+alpha(n-1)` `therefore alpha(n-1)=hati-1` `therefore alpha=(i-1)/((n-1))` This is the relation between degree of dissociation and van't Hoff factor i. |
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