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Van’t Hoff factor (i) for the centimolal solution of ${{\text{K}}_{\text{3}}}\text{ }\!\![\!\!\text{ Fe(CN}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }$ is 3.333. What is it’s percentage dissociation?
Van’t Hoff factor (i) for the centimolal solution of ${{\text{K}}_{\text{3}}}\text{ }\!\![\!\!\text{ Fe(CN}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }$ is 3.333. What is it’s percentage dissociation?
Chemistry | Chemistry | MHTCET
Van’t Hoff factor (i) for the centimolal solution of ${{\text{K}}_{\text{3}}}\text{ }\!\![\!\!\text{ Fe(CN}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }$ is 3.333. What is it’s percentage dissociation?
  1. 80%
  2. 70%
  3. 33.33%
  4. 77.7%

Solution: 77.7%

$ {{K}_{3}}[Fe{{(CN)}_{6}}]\to 3{{K}^{+}}+{{[Fe{{(CN)}_{6}}]}^{3-}} $

$ We\,get: $

$ i=\alpha n+(1-\alpha )=1+\alpha (n-1)=1+\alpha (4-1)~ $

$ Van’t\text{ }Hoff\text{ }factor~(i)=3.333 $

$ n=~Total\text{ }moles\text{ }formed\text{ }after\text{ }dissociation.\,Let\text{ }degree\text{ }of\text{ }dissociation\text{ }be~\alpha . $

$ So,\,we\,have: $

$ 3.333=1+\alpha (4-1)~ $

$ \alpha (4-1)=3.333-1~ $

$ \alpha (4-1)=2.333 $

$ 3\alpha =2.333 $

$ \alpha =2.333/3=0.777~ $

$ Percent\text{ }dissociation~=\alpha \times 100=77.7 $

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