1.22 g of an organic acid is separately dissolved in 100 g of benzene (Kb = 2.6 K kg mol^-1) and 100

Q: Why do we assume 100% dimerisation in benzene? The question doesn't explicitly state it.

In JEE problems, if a solute is known to associate or dissociate and no degree of association/dissociation ($\alpha$) is provided, it is a standard convention to assume 100% completion. For dimerisation, this means $\alpha = 1$, leading to $i = 1/2$.

Q: Why does the organic acid dimerise in benzene but not in acetone?

Benzene is a non-polar solvent. In the absence of competing interactions, organic acids (like carboxylic acids) form strong intermolecular hydrogen bonds with each other, creating dimers. Acetone, however, is a polar aprotic solvent that can form hydrogen bonds with the acid molecules, preventing them from bonding with each other.

Q: What is the general formula for the van't Hoff factor ($i$) during association?

The general formula is $i = 1 - \alpha + \frac{\alpha}{n}$, where $\alpha$ is the degree of association and $n$ is the number of molecules associating into one unit. For 100% dimerisation, $\alpha = 1$ and $n = 2$, which gives $i = 1 - 1 + \frac{1}{2} = 0.5$.

Q: Could I have solved this by finding the molar mass first?

Absolutely! You could use the acetone data to find the molar mass ($M_w = 122 \text{ g/mol}$), and then use that $M_w$ in the benzene equation to find $\Delta T_b$. However, the ratio method is much faster and less prone to calculation errors.

Q: What if the masses of the solvents were different?

If the masses of the solvents were different, the molality ($m$) would not cancel out when taking the ratio. In that case, you would either have to calculate the molar mass first, or expand the molality term in the ratio equation and plug in the different solvent masses.

Question

1.22 g of an organic acid is separately dissolved in 100 g of benzene ($K_b = 2.6 	ext{ K kg mol}^{-1}$) and 100 g of acetone ($K_b = 1.7 	ext{ K kg mol}^{-1}$). The acid is known to dimerise in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by $0.17^\circ	ext{C}$. The increase in boiling point of solution in benzene in $^\circ	ext{C}$ is $x 	imes 10^{-2}$. The value of $x$ is ........ (Nearest integer)
[Atomic mass : C = 12.0, H = 1.0, O= 16.0]

Sigma Percentile · Accepted Answer

### The Tale of Two Solvents

Imagine you are a chemist in a laboratory, holding a vial containing exactly $1.22 	ext{ g}$ of an unknown organic acid. You decide to run an experiment by dissolving this acid into two completely different environments: $100 	ext{ g}$ of Acetone and $100 	ext{ g}$ of Benzene.

What happens next is a beautiful demonstration of how a solvent's nature dictates the behavior of the molecules within it. In the polar environment of acetone, the organic acid molecules are perfectly content floating around individually. They remain as **monomers**. However, when dropped into the non-polar world of benzene, the acid molecules seek each other out, forming hydrogen bonds to create pairs. They **dimerise**.

### The Master Equation: Elevation in Boiling Point

Whenever we add a non-volatile solute to a solvent, the boiling point of the solution increases. This phenomenon is governed by the equation:

$$\Delta T_b = i \cdot K_b \cdot m$$

Here, $i$ is the van't Hoff factor, $K_b$ is the boiling point elevation constant, and $m$ is the molality of the solution.

Now, let's look at our setup with a strategic eye. In both beakers, we have the exact same mass of solute ($1.22 	ext{ g}$) and the exact same mass of solvent ($100 	ext{ g}$). This means the **molality ($m$) is perfectly identical** for both solutions! Therefore, the elevation in boiling point is directly proportional to the product of the van't Hoff factor and the $K_b$ value.

### Analyzing the Acetone Solution

In the acetone beaker, the acid remains a monomer. It neither associates nor dissociates.

This gives us a van't Hoff factor of $i_{	ext{ace}} = 1$. We are given the constant $K_{b(	ext{ace})} = 1.7 	ext{ K kg mol}^{-1}$, and the observed elevation in boiling point is $\Delta T_{b(	ext{ace})} = 0.17^\circ	ext{C}$.

### Analyzing the Benzene Solution

Now, let's shift our focus to the benzene beaker. Here, the acid undergoes dimerisation ($2	ext{A} ightleftharpoons 	ext{A}_2$). Since no degree of association is provided, we must assume **100% association**.

When two molecules combine to form one, the effective number of particles is halved. Thus, our van't Hoff factor becomes $i_{	ext{ben}} = \frac{1}{2} = 0.5$. The constant for benzene is $K_{b(	ext{ben})} = 2.6 	ext{ K kg mol}^{-1}$.

### The Ninja Technique: Taking the Ratio

Many students would use the acetone data to painstakingly calculate the molar mass of the organic acid, and then plug that mass into the benzene equation. While correct, it is time-consuming. Let's use a more elegant approach: **The Ratio Method**.

By dividing the equation for benzene by the equation for acetone, the constant molality ($m$) completely cancels out:

$$\frac{\Delta T_{b(	ext{ben})}}{\Delta T_{b(	ext{ace})}} = \frac{i_{	ext{ben}} \cdot K_{b(	ext{ben})}}{i_{	ext{ace}} \cdot K_{b(	ext{ace})}}$$

Let's substitute our known values into this streamlined equation:

$$\frac{\Delta T_{b(	ext{ben})}}{0.17} = \frac{0.5 \cdot 2.6}{1 \cdot 1.7}$$

### The Final Calculation

Now, it is just a matter of simple arithmetic. Multiplying $0.5$ by $2.6$ gives us $1.3$.

$$\frac{\Delta T_{b(	ext{ben})}}{0.17} = \frac{1.3}{1.7}$$

Cross-multiplying the $0.17$ to the other side:

$$\Delta T_{b(	ext{ben})} = 0.17 \cdot \frac{1.3}{1.7} = 0.13^\circ	ext{C}$$

The problem asks for the answer in the format of $x 	imes 10^{-2} ^\circ	ext{C}$. We can easily rewrite our result:

$$0.13^\circ	ext{C} = 13 	imes 10^{-2} ^\circ	ext{C}$$

Comparing this to the given format, we find that **$x = 13$**. A beautiful, clean integer derived from a deep understanding of physical chemistry!

Enter Numerical Value:

Visualized Solution

$The Dual Solvent Setup$

$Elevation in Boiling Point$

$Case 1: Acetone Solution$

$Case 2: Benzene Solution$

$Taking the Ratio$

$Calculating Δ T_{b (ben)}$

$Finding x$

The Sigma Insight: Abnormal Molecular Mass and Distribution Law

The Tale of Two Solvents

The Master Equation

Analyzing the Acetone Solution

Analyzing the Benzene Solution

The Ninja Technique

The Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Dual Solvent Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elevation in Boiling Point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 1: Acetone Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: Benzene Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Taking the Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ΔTb(ben)​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding x