Question

an electrolyte ab has degree of dissociation 60%. what will be the value of van't hoff factor for the electrolyte?​

Answer 1

hey there !

we know that ,

α= i - 1 / n- 1

where ,  α =  degree of dissociation and i = van't hoff factor

60/100 = i -1 /2-1   (n= 2 for ab)

0.6  = i-1

1.6 = i

So , the value of van't hoff factor for the electrolyte is 1.6

# hope it helps :)

Answer 2

Answer:

2

Explanation:

The van’t Hoff component is sincerely only a mathematical component that scales the combined or label attention of a solute in order that it suits the real or general attention of all species generated with the aid of using that solute after dissolution.

Solutes commonly are available in 3 sorts that we're involved with: non-electrolytes, vulnerable electrolytes, and sturdy electrolytes. Each of those may be effortlessly categorised with the aid of using the manner wherein they dissociate and the ions that they shape or do now no longer shape because the case can also additionally be. These solutes dissolve and in part dissociate or ionize to yield an answer of the discern species plus the ions which can be generated after ionization. They normally handiest ionize in small chances together with 1 to 5%. However, any percentage this is much less than 100% is taken into consideration a “vulnerable” electrolyte. The van’t Hoff component for those solutes may be any actual quantity among 1 and 2.

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