When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found

Q: Why does the freezing point decrease when we add benzoic acid?

This is due to freezing point depression, a colligative property. Adding a solute disrupts the formation of the solid solvent lattice, requiring a lower temperature to freeze the solution compared to the pure solvent.

Q: What exactly does '100% association' mean?

It means that every single molecule of benzoic acid you put into the water participates in forming the associated complex. None are left floating around as individual, unassociated molecules. Mathematically, this means the degree of association, $\alpha$, is exactly 1.

Q: Why is the van't Hoff factor ($i$) less than 1 here?

The van't Hoff factor represents the ratio of actual particles in solution to the theoretical number of particles. Because the molecules are clumping together (associating), there are fewer total particles floating in the water than we initially added. Hence, $i < 1$.

Q: How did we get the molar mass of benzoic acid to be 122?

The chemical formula for benzoic acid is $\text{C}_6\text{H}_5\text{COOH}$. By adding the atomic masses of 7 Carbon atoms ($7 \times 12$), 6 Hydrogen atoms ($6 \times 1$), and 2 Oxygen atoms ($2 \times 16$), we get $84 + 6 + 32 = 122 \text{ g/mol}$.

Q: What if the solvent was benzene instead of water?

Benzoic acid is actually famous for dimerizing in benzene due to hydrogen bonding! The mathematical process would be identical, but the $K_f$ value for benzene would be different from water's $1.86$.

Question

When $12.2	ext{ g}$ of benzoic acid is dissolved in $100	ext{ g}$ of water, the freezing point of solution was found to be $- 0.93^\circ	ext{C}$ ($K_f(H_2O) = 1.86 	ext{ K kg mol}^{-1}$). The number ($n$) of benzoic acid molecules associated (assuming 100% association) is ……… .

Sigma Percentile · Accepted Answer

### The Setup: A Chilly Conundrum

Imagine you are in a chemistry lab, and you have a beaker filled with exactly $100	ext{ g}$ of pure water. You carefully weigh out $12.2	ext{ g}$ of benzoic acid ($	ext{C}_6	ext{H}_5	ext{COOH}$) and dissolve it into the water.

Normally, you could predict the new freezing point of this solution using standard colligative property formulas. However, when you measure it, the freezing point has dropped to $-0.93^\circ	ext{C}$. This is our $\Delta T_f$. Something interesting is happening at the molecular level: the benzoic acid molecules are not just floating around independently; they are teaming up! Our mission is to find out how many molecules ($n$) are joining together to form a single associated complex.

### The Master Equation: Colligative Properties

To solve this mystery, we need our trusty colligative property equation for freezing point depression, modified by the van't Hoff factor ($i$):

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

Here, $K_f$ is the molal depression constant for water ($1.86 	ext{ K kg mol}^{-1}$), and $m$ is the molality of the solution. The van't Hoff factor, $i$, is the key to unlocking the association. It tells us the ratio of the actual number of particles in solution to the number of particles we *thought* we put in.

### The Reveal: Where Did the Molecules Go?

Before we can find $i$, we need the molality ($m$). Molality is the moles of solute per kilogram of solvent. Let's calculate the molar mass of benzoic acid ($M_B$):

$$M_B = (7 	imes 12) + (6 	imes 1) + (2 	imes 16) = 122 	ext{ g mol}^{-1}$$

Now, let's plug our values into the molality formula:

$$m = \frac{w_B 	imes 1000}{M_B 	imes w_A} = \frac{12.2 	imes 1000}{122 	imes 100} = 1 	ext{ mol kg}^{-1}$$

With a molality of exactly $1 	ext{ mol kg}^{-1}$, we can substitute everything back into our freezing point equation:

$$0.93 = i 	imes 1.86 	imes 1$$

Solving for $i$, we get:

$$i = \frac{0.93}{1.86} = 0.5$$

An $i$ value of $0.5$ means that for every $100$ molecules we put in, they act like only $50$ particles! They are definitely pairing up.

### The Final Calculation: Solving for $n$

The problem states that there is **100% association**, which means the degree of association ($\alpha$) is $1$. The general formula relating the van't Hoff factor to association is:

$$i = 1 - \alpha + \frac{\alpha}{n}$$

Substituting $\alpha = 1$:

$$i = 1 - 1 + \frac{1}{n} = \frac{1}{n}$$

Since we already found that $i = 0.5$, we can equate the two:

$$\frac{1}{n} = 0.5$$

$$n = 2$$

**Conclusion:** The benzoic acid molecules associate in pairs, forming dimers in the solution. The final answer is **2**.

Enter Numerical Value:

Visualized Solution

Visual Anchor: The Setup

Logic Bridge: The Master Equation

Raw Setup: Molar Mass & Molality

Atomic Compute: Finding 'i'

Logic Bridge: Degree of Association

Final Answer: Solving for 'n'

The Way Forward: Partial Association

The Sigma Insight: Abnormal Molecular Mass and Distribution Law

The Setup

The Master Equation

The Reveal

The Final Calculation

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