Question

If the van’t Hoff-factor for $0.1MBa(N{O}_3{)}_2$ solution is $2.74$, the degree of dissociation is

• A. $0.87$
• B. $0.74$
• C. $0.91$
• D. $87$

Answer 1

Answer: (A)

Given,

Molarity $=01M$ vant Hoff factor $(i)=2.74$

Since, $i>1$, it means solute is undergoing dissociation.

$Ba{\left(N{O}_3\right)}_2\rightleftharpoons B{a}^{2+}+2N{O}_3^{-}$

Number of particles dissociated $(n)=3$

Now, $\alpha$ (degree of dissociation) $=\frac{(i-1)}{(n-1)}$

$=\frac{(2.74-1)}{(3-1)}$

$=0.87$