Question

the van't hoff's factor of 0.1M Ba(NO2)2 solution is 2.74. the degree of dissociation is -

Answer 1

$\bold{ans : 0.87}$

actually, van't Hoff's factor for any reaction ,
$A_n---->nA \\ 1 - \alpha \: \: \: \: \: \: \: \: \: \: \: \: \: \: n \alpha$
then, van't Hoff factor ( i ) = 1 + (n -1)α
where α is the degree of dissociation .
similarly,

$Ba(NO_3)_2---->Ba^+ +2NO_3^- \\ 1 - \alpha \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \alpha \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2\alpha$

now , van't Hoff' factor { i } = 1 - α + α + 2α
2.74 = 1 + 2α
1.74 = 2α
α = 0.87

Answer 2

Chemical Equation involved :

Ba(NO3)2 ------> Ba²⁺ + 2NO3⁻

At the start 
1 mole                 0              0

At equilibrium
(moles 1-α)         α             2α

total number of ions(mole)=1-α+α+2α

=1+2α
1+2α=i

α=i-1/2
=2.74-1/2
=1.74/2
=0.87
=87%