Question:
$ 2 \,g $ of benzoic acid $ (C_6H_6COOH) $ is dissolved in $ 25\, g $ of benzene. The observed molar mass of benzoic acid is found to be $ 241.98 $ . What is the percentage association of acid if it forms in solution?
$ 2 \,g $ of benzoic acid $ (C_6H_6COOH) $ is dissolved in $ 25\, g $ of benzene. The observed molar mass of benzoic acid is found to be $ 241.98 $ . What is the percentage association of acid if it forms in solution?
Updated On: Jul 28, 2023
- $ 85.2\% $
- $ 89.2\% $
- $ 95.2\% $
- $ 99.2\% $
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The Correct Option is D
Solution and Explanation
Given $W_A$, benzoic acid $= 2 \,g$
$W_B$, Benzene $= 25 \,g$
$K_f = 4.9 \,K \,kg\, mol^{-1}$
$T_f = 1.62 \,K$
$M_B =241.98$
$2C_6H_5COOH {<=>} (C_6H_5COOH)_2$
Let, $x =$ degree of association of solute
$\therefore (1 - x) =$ moles of benzene left in unassociated form
$\frac{x}{2}=$ moles of benzoic acid in equilibrium
Total number of moles at equilibrium $= 1 - x + \frac{x}{2}$
van,t Hoff factor, $ i = 1 - \frac{x}{2}$
$i = \frac{\text{Total number of moles of particles after association }}{\text{Number of moles of particle before association}}$
$i = \frac{122\,g\,mol}{241.98\,g\,mol}$
or, $ 1 - \frac{x}{2} = \frac{122}{241.98}$
$\therefore \frac{x}{2} = 1 - \frac{122}{241.98} $
$\frac{x}{2} = 0.49$
$ x = 2\times 0.49$
$ x = 0.9916$
$\therefore$ The degree of association of benzoic acid in benzene is $99.2\%$.
$W_B$, Benzene $= 25 \,g$
$K_f = 4.9 \,K \,kg\, mol^{-1}$
$T_f = 1.62 \,K$
$M_B =241.98$
$2C_6H_5COOH {<=>} (C_6H_5COOH)_2$
Let, $x =$ degree of association of solute
$\therefore (1 - x) =$ moles of benzene left in unassociated form
$\frac{x}{2}=$ moles of benzoic acid in equilibrium
Total number of moles at equilibrium $= 1 - x + \frac{x}{2}$
van,t Hoff factor, $ i = 1 - \frac{x}{2}$
$i = \frac{\text{Total number of moles of particles after association }}{\text{Number of moles of particle before association}}$
$i = \frac{122\,g\,mol}{241.98\,g\,mol}$
or, $ 1 - \frac{x}{2} = \frac{122}{241.98}$
$\therefore \frac{x}{2} = 1 - \frac{122}{241.98} $
$\frac{x}{2} = 0.49$
$ x = 2\times 0.49$
$ x = 0.9916$
$\therefore$ The degree of association of benzoic acid in benzene is $99.2\%$.
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Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.