The von't Hoff factor for 0.1M Ba(NO_{3})_{2} solution is 2.74. The degree of dissociation is

  • Option 1)

    91.3%

  • Option 2)

    74 %

  • Option 3)

    87%

  • Option 4)

    100%

 

Answers (1)
D Divya Saini

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

 

 i= 1+\alpha (n-1)\: where, \:\alpha = degree \: of\: dissociation

\Rightarrow 2.74 -1 =\alpha (2)

\Rightarrow\alpha = 0.87


Option 1)

91.3%

Incorrect

Option 2)

74 %

Incorrect

Option 3)

87%

Correct

Option 4)

100%

Incorrect

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