Question

The van't Hoff factor of $BaC{l_2}$​ at $0.01M$ concentration is $1.98$. The percentage of dissociation of $BaC{l_2}$​ at this concentration is:
A. $49$
B. $69$
C. $89$
D. $100$

Answer 1

Hint: Some chemical compounds dissociate or associate in solution when utilised as solutes. Their molar mass and other colligative characteristics change as a result. The Van't Hoff Factor can account for this. It was given the name Jacobus Henricus Van't Hoff, Jr. after the Dutch physical chemist.

Formula Used: van’t Hoff factor which is represented as:
$i = \dfrac{{Number\,of\,particles\,after\,dissociation}}{{Number\,of\,particles\,before\,dissociation}}$

Complete step-by-step answer:In the question, we have given the van’t Hoff factor of $BaC{l_2}$ at $0.01M$ concentration is $1.98$,
We can consider the dissociation of $BaC{l_2}$, we have
$BaC{l_2} \rightleftharpoons B{a^{2 + }} + 2C{l^ - }$
From the given information, the concentration is $0.01M$ at the time $t = 0$
Let assume the number of particles be $x$, then at the time $t$:
$(0.01 - x) \rightleftharpoons xM + 2xM$
Now, use the formula of van’t Hoff factor which is represented as:
$i = \dfrac{{Number\,of\,particles\,after\,dissociation}}{{Number\,of\,particles\,before\,dissociation}}$
Substitute the values in the above formula, then we have:
$i = \dfrac{{(0.01 - x) + x + 2x}}{{0.01}} \\$
 $\Rightarrow i = \dfrac{{0.01 + 2x}}{{0.01}} \\$
As we know that the van't Hoff factor of $BaC{l_2}$​ at $0.01M$concentration is $1.98$, i.e., $i = 1.98$
$1.98 = \dfrac{{0.01 + 2x}}{{0.01}} \\$
$\Rightarrow 1.98 \times 0.01 = 0.01 + 2x \\$
 $\Rightarrow x = \dfrac{{0.0198 - 0.01}}{2} \\$
 $\Rightarrow x = 0.0049 \\$
When introduced in water, strong acids and bases totally dissociate. In other words, the full base $[BOH]$ or acid $[HA]$ splits into ions.

On the other hand, the weak acids or the bases do not entirely dissociate in an aqueous solution. The degree to which they separate is indicated by the dissociation constants ${K_a}$ and ${K_b}$:
$\begin{array}{*{20}{l}}
{{K_a}\; = {\text{ }}\left( {\left[ {{H^ + }} \right]\left[ {{A^-}} \right]} \right){\text{ }} \div {\text{ }}\left[ {HA} \right]} \\
  {{K_b}\; = {\text{ }}(\left[ {{B^ + }} \right]\left[ {O{H^-}} \right]){\text{ }} \div {\text{ }}\left[ {BOH} \right]}
\end{array}$
Therefore, the dissociation percentage is:
$\alpha = \dfrac{x}{{0.01}} \times 100 \\$
 $\Rightarrow \dfrac{{0.0049}}{{0.01}} \times 100 \\$
  $\Rightarrow 49\% \\$

Option ‘A’ is correct

Note:When the amount of solute increases (the association value of $i$ is more than one), its colligative property also grows. The amount of solute reduces, the colligative property reduces, when the association value of $i$ is less than one.