Question

The van't Hoff factor for $0.1M$ barium nitrate is $2.74$. The percentage of dissociation of barium nitrate is:

• A. $91.3\%$
• B. $87\%$
• C. $100\%$
• D. $74\%$

Answer 1

The correct option is B $87\%$

In $Ba(N{O}_3{)}_2$

$n=3$ (no. of ions)

Now,
$i=1+\alpha (n-1)$

$2.74=1+\alpha (3-1)$

$\alpha =\frac{1.74}{2}$

$\alpha =0.87$

$\therefore \alpha$% $=0.87\times 100$

$\therefore \alpha \%=87$%