Which one of the following electrolytes has the same value of van't Hoff's factor (i) as that of Al2(SO4)(if all are 100% ionised) ?

  • Option 1)

    Al(NO3)3

  • Option 2)

    K4[Fe(CN)6]

  • Option 3)

    K2SO4

  • Option 4)

    K3[Fe(CN)6]

 

Answers (1)

As we learned in concept:

 

Vant Hoff factor for dissociation -

i= 1+(n-1)\alpha

Where

n is the no. of dissociated particles

\alpha = degree of association
 

- wherein

NaC l \: \: \: n = 2

CaCl_{2} \: \: \: n = 3

K_{4}[F(CN_{6})]\: \: \: \: \: n=5

 

 Let's check Van't Haff factor for each option.

 

Al(No3)3 -> Al3+ + 3(No3)-, i = 4 

K4[ Fe (CN)6] -> 4K+ + [ Fe (CN)6]4-, i = 5

K2So4 -> 2K+ + So42-, i=3

K4[Fe (CN)6] -> 3K+ + [Fe (CN)6]3-, i = 4

Al2(So4)3 -> 2Al3+ + 3 So42-, i = 5

Clearly, answer is K4[Fe(CN)6]


Option 1)

Al(NO3)3

this is the incorrect option

Option 2)

K4[Fe(CN)6]

this is the correct option

Option 3)

K2SO4

this is the incorrect option

Option 4)

K3[Fe(CN)6]

this is the incorrect option

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