Explanation

(b)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Ba(NO_{3})_{2}\ \ \ \ \ \ \rightleftharpoons \ \ \ Ba^{2 +}\ \ \ + \ \ \ \ \ \ 2NO_{3}^{-}$
At t=0 0.1 M 0 0
At equilibrium (0.1-x)M xM 2xM
i=$\frac{(0.1 - x) + x + 2x}{0.1}$
2.74=$\frac{0.1 + 2x}{0.1}$
0.1+2x=0.274
2x=0.274-0.1=0.174
x=$\frac{0.174}{2}$=0.087
$\therefore$ Degree of dissociation =$\frac{0.087}{0.1} \times 100 = 87\%\ \$