The value of van't Hoff factor for $0.1$ M $B a (N O_{3})_{2}$ solution is $2.74$ . The degree of dissociation is:

100 %

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92 %

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87 %

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74 %

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Solution

The correct option is C $87 %$
$B a (N O_{3})_{2}$ dissociate as:

Initial at equilibrium,
$B a (N O_{3})_{2} 0.1 M (0.1 - x) M ⇋ B a_{0}^{2 +} x a M + 2 N O_{3}^{-} 0 2 x 2 a M$ .
Thus, n $= 3$
$i = n α + (1 - α) = 3 α + 1 - α$ (per mole)
Van't Hoff factor $i = 1 + 2 α \Rightarrow 2.74 = 1 + 2 α$
$2 α = 1.74$
$α = 0.87$
$α (%) = 87 %$ .

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