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The Van’t Hoff factor for $0\cdot 1M$$Ba{{\left( N{{O}_{3}} \right)}_{2}}$ solution is $\text{2}\cdot \text{74}$ The degree of dissociation is:A.$91.3%$ B.$87%$ C.$100%$ D.$74%$

Last updated date: 12th Sep 2024
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Hint: The Van’t Hoff factor is used to determine the degree of dissociation. It is a measure of deviation from ideal behavior. With the increasing concentration of solute, the Van’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution.

Formula used: $i=1+2\alpha$ where i= Van’t Hoff Factor
$\alpha$= degree of dissociation

The solution It is defined as the ratio of the concentration of particles formed when a substance is dissolved to the concentration of the substance by mass. The extent to which a substance associates or dissociates in a solution is described by Van’t Hoff Factor. This Factor is represented by ‘$i$ ’ and is named after Dutch physical chemist Jacobus Henricus Van’t Hoff.
For an Electrolytic solution, the value of $i$ is generally higher than the predicted value (due to pairing of ions).
Association: It is the joining of two or more particles to form one entity ($i$ is less than 1).
Dissociation: It refers to the splitting of a molecule into multiple iconic entities ($i$ is greater than 1).
Given compound, $Ba{{(N{{O}_{3}})}_{2}}\text{ dissociates into B}{{\text{a}}^{2+}}\text{ and 2N}{{\text{O}}^{-3}}$ . Using,
$i=1+2\alpha$
$\Rightarrow 2\cdot 74=1+2\alpha$
$\Rightarrow 2\cdot 74-1=2\alpha$
$\Rightarrow 1.74=2\alpha$
$\Rightarrow \alpha =\dfrac{1\cdot 74}{2}=0\cdot 87$
$\Rightarrow \alpha =87%$

So, the correct answer is “Option B”.

i.e. $i=\dfrac{measured\text{ value of concentration of particles}}{calculated\text{ value of concentration from its mass}}$