Molecules of benzoic acid (C6H5COOH) dimerise in benzene. 'w' g of the acid dissolved in 30 g of ben

Q: Why does benzoic acid dimerize in benzene but not in water?

Benzene is a non-polar solvent. In the absence of a polar solvent to interact with, the polar carboxylic acid groups of benzoic acid molecules hydrogen-bond with each other, forming stable dimers. In water, the molecules would hydrogen-bond with the water itself and even dissociate into ions.

Q: Why is the formula for 'i' written as $1 - \alpha + \frac{\alpha}{2}$?

If you start with 1 mole, $\alpha$ moles react. So you are left with $1 - \alpha$ moles of the original monomer. Since 2 monomers form 1 dimer, the $\alpha$ moles that reacted will form $\frac{\alpha}{2}$ moles of the dimer. The total particles are the sum of these two.

Q: What if the percentage association was 100%?

If association was 100%, then $\alpha = 1$. The van't Hoff factor would be $i = 1 - 1 + \frac{1}{2} = 0.5$. This means exactly half the number of particles are present compared to what was dissolved.

Q: Why do we multiply by 1000 in the molality formula?

Molality is defined as moles of solute per **kilogram** of solvent. Since the mass of the solvent (benzene) is given in grams (30 g), we divide it by 1000 to convert it to kilograms. Mathematically, this 1000 moves to the numerator.

Q: Can we use this same 'i' value if we were calculating boiling point elevation?

Yes! The van't Hoff factor $i$ depends purely on the state of the solute in the solvent (how much it associates or dissociates). It applies equally to all colligative properties: freezing point depression, boiling point elevation, and osmotic pressure.

Question

Molecules of benzoic acid ($	ext{C}_6	ext{H}_5	ext{COOH}$) dimerise in benzene. '$w$' g of the acid dissolved in $30	ext{ g}$ of benzene shows a depression in freezing point equal to $2	ext{ K}$. If the percentage association of the acid to form dimer in the solution is $80$, then $w$ is
(Given that $K_f = 5	ext{ K kg mol}^{-1}$, molar mass of benzoic acid $= 122	ext{ g mol}^{-1}$)

Options:

(A) 1.8 g

(B) 1.0 g

(C) 2.4 g

(D) 1.5 g

Sigma Percentile · Accepted Answer

### The Mystery of the Disappearing Molecules

Imagine you are a chemist looking at a beaker filled with $30	ext{ g}$ of benzene. You carefully weigh out an unknown amount, let's call it $w	ext{ g}$, of benzoic acid ($	ext{C}_6	ext{H}_5	ext{COOH}$) and dissolve it. You expect a certain depression in the freezing point based on the number of molecules you added. But nature has a trick up its sleeve!

In non-polar solvents like benzene, carboxylic acids like benzoic acid don't like to stay alone. Thanks to intermolecular hydrogen bonding, they pair up to form **dimers**.

$$2	ext{C}_6	ext{H}_5	ext{COOH} ightleftharpoons (	ext{C}_6	ext{H}_5	ext{COOH})_2$$

This means the actual number of particles floating in the solution is *less* than what you initially put in. Since colligative properties like freezing point depression depend strictly on the *number* of particles, the observed depression will be less than the theoretical one.

### Unmasking the van't Hoff Factor

To account for this molecular pairing, we use the **van't Hoff factor ($i$)**. It is the ratio of the actual number of particles in solution to the number of particles initially dissolved.

Let's set up the equilibrium. If we start with $1$ mole of benzoic acid and its degree of association is $\alpha$:
- Moles of unassociated benzoic acid remaining $= 1 - \alpha$
- Moles of dimer formed $= \frac{\alpha}{2}$

The problem states that the percentage association is $80\%$, which means $\alpha = 0.8$.

Let's calculate the total moles at equilibrium:
$$	ext{Total moles} = (1 - 0.8) + \frac{0.8}{2}$$
$$	ext{Total moles} = 0.2 + 0.4 = 0.6$$

Therefore, our van't Hoff factor is:
$$i = \frac{0.6}{1} = 0.6$$

An $i$ value less than $1$ perfectly confirms that our particles are associating!

### The Freezing Point Equation

Now, we bring in the master equation for the depression in freezing point:

$$\Delta T_f = i 	imes K_f 	imes m$$

We know that molality ($m$) is the moles of solute per kilogram of solvent. Expanding the molality term, we get:

$$\Delta T_f = i 	imes K_f 	imes \frac{w_{	ext{solute}} 	imes 1000}{Mw_{	ext{solute}} 	imes w_{	ext{solvent}}}$$

We are given all the necessary parameters:
- $\Delta T_f = 2	ext{ K}$
- $K_f = 5	ext{ K kg mol}^{-1}$
- $Mw_{	ext{solute}} = 122	ext{ g mol}^{-1}$
- $w_{	ext{solvent}} = 30	ext{ g}$

### The Final Calculation

Let's substitute our known values into the expanded equation:

$$2 = 0.6 	imes 5 	imes \frac{w 	imes 1000}{122 	imes 30}$$

Now, it's just a matter of careful algebra. Let's simplify the numerator and the denominator:

$$2 = \frac{3 	imes w 	imes 1000}{3660}$$
$$2 = \frac{3000 w}{3660}$$

Rearranging to solve for our unknown mass $w$:

$$w = \frac{2 	imes 3660}{3000}$$
$$w = \frac{7320}{3000} = 2.44	ext{ g}$$

Looking at our options, $2.4	ext{ g}$ is the closest match. The beauty of this problem lies in realizing that without the van't Hoff factor, our calculated mass would have been completely wrong. Always respect the solvent-solute dynamics!

Select Answer:

Visualized Solution

$Visualizing the Solution$

$The Dimerization Reaction$

$Setting up van’t Hoff Factor (i)$

$Calculating Total Moles$

$Finalizing i$

$Depression in Freezing Point$

$Substituting the Values$

$Simplifying the Equation$

$Solving for w$

$The Way Forward$

The Sigma Insight: Abnormal Molecular Mass and Distribution Law

The Mystery of the Disappearing Molecules

Unmasking the van't Hoff Factor

The Freezing Point Equation

The Final Calculation

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Select Answer:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up van’t Hoff Factor (i)

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

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

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

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

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

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

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