MX2 dissociates into M2+ and X- ions in an aqueous solution, with a degree of dissociation (alpha) o

Q: What does the ratio of observed to theoretical depression of freezing point represent?

This ratio is the definition of the van 't Hoff factor ($i$). It accounts for the change in the number of particles in solution due to dissociation or association.

Q: How do we find the value of $n$?

$n$ is the total number of ions produced when one formula unit of the solute dissociates completely. For $\text{MX}_2 \rightarrow \text{M}^{2+} + 2\text{X}^{-}$, $n = 1 + 2 = 3$.

Q: What if the degree of dissociation ($\alpha$) was 1?

If $\alpha = 1$ (100% dissociation), the formula $i = 1 + \alpha(n - 1)$ simplifies to $i = n$. In this case, $i$ would be 3.

Q: Why is the theoretical depression of freezing point calculated without ionic dissociation?

The theoretical value assumes the solute does not dissociate or associate, meaning 1 mole of solute provides exactly 1 mole of particles in solution.

Q: Can this formula be used for association?

Yes, but for association, the formula is $i = 1 + \alpha(1/n - 1)$, where $n$ is the number of molecules associating into a single larger molecule.

Question

$	ext{MX}_2$ dissociates into $	ext{M}^{2+}$ and $	ext{X}^{-}$ ions in an aqueous solution, with a degree of dissociation ($\alpha$) of $0.5$. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is

Sigma Percentile · Accepted Answer

### The Mystery of the Freezing Point

Imagine you're trying to freeze water, but you've added a pinch of salt. You probably know that adding a solute lowers the freezing point of the solvent—a phenomenon known as the **depression of freezing point**. But what happens when that solute breaks apart into multiple pieces once it hits the water?

This is where the concept of the **van 't Hoff factor ($i$)** comes into play. The van 't Hoff factor is essentially a multiplier. It tells us how many particles we actually get in the solution compared to how many formula units we dissolved.

### Decoding the Question

The problem asks for the ratio of the **observed depression of freezing point** to the **depression of freezing point in the absence of ionic dissociation** (which is the theoretical value).

Mathematically, this ratio is the very definition of the van 't Hoff factor:

$$ i = \frac{(\Delta T_f)_{	ext{observed}}}{(\Delta T_f)_{	ext{theoretical}}} $$

So, our mission is simply to calculate $i$.

### The Dissociation Dance

We are given a generic salt, $	ext{MX}_2$. When it dissolves, it undergoes dissociation:

$$ 	ext{MX}_2 ightleftharpoons 	ext{M}^{2+} + 2	ext{X}^{-} $$

If one molecule of $	ext{MX}_2$ were to break apart completely, it would yield 1 $	ext{M}^{2+}$ ion and 2 $	ext{X}^{-}$ ions, giving a total of $n = 3$ ions.

However, the problem states that the **degree of dissociation ($\alpha$)** is $0.5$. This means only 50% of the $	ext{MX}_2$ molecules actually break apart.

### The Master Equation

To find the effective number of particles (the van 't Hoff factor) when dissociation is partial, we use the master formula:

$$ i = 1 + \alpha(n - 1) $$

Let's break down why this works. You start with 1 mole of $	ext{MX}_2$. A fraction $\alpha$ dissociates, leaving $(1 - \alpha)$ moles of intact $	ext{MX}_2$. The dissociated portion turns into $n\alpha$ moles of ions. The total number of particles is $(1 - \alpha) + n\alpha = 1 + \alpha(n - 1)$.

### The Final Calculation

Now, we just plug in our values:
- $\alpha = 0.5$
- $n = 3$

$$ i = 1 + 0.5(3 - 1) $$

$$ i = 1 + 0.5(2) $$

$$ i = 1 + 1 = 2 $$

The van 't Hoff factor is exactly **2**. This means that, on average, every formula unit of $	ext{MX}_2$ we add results in 2 particles in the solution. Consequently, the observed freezing point depression will be exactly twice the theoretical value we would expect if no dissociation occurred.

Enter Numerical Value:

Visualized Solution

$Dissociation Reaction$

$van ’t Hoff Factor (i)$

$Formula for i$

$Substitution$

$Calculation$

$Final Answer$

The Sigma Insight: Abnormal Molecular Mass and Distribution Law

The Mystery of the Freezing Point

Decoding the Question

The Dissociation Dance

The Master Equation

The Final Calculation

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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; } Dissociation Reaction

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

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

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

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

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