An aqueous solution is prepared by dissolving 0.1 mol of an ionic salt in 1.8 kg of water at 35 °C.

Q: Why did we use $(P^\circ - P_s) / P_s = n/N$ instead of $(P^\circ - P_s) / P^\circ = n/(n+N)$?

Both formulas are mathematically identical! The form with $P_s$ in the denominator and just $N$ on the right side is derived by taking the reciprocal of the standard formula and subtracting 1. It makes calculations significantly faster by avoiding the $(n+N)$ term in the denominator.

Q: How did we get the expression $0.1(1 - 0.9)$ for the undissociated salt?

If you start with $C$ moles and the degree of dissociation is $\alpha$, the amount that dissociates is $C\alpha$. The amount left behind is $C - C\alpha$, which factors to $C(1 - \alpha)$. Here, $C = 0.1$ and $\alpha = 0.9$.

Q: Why do we multiply by 'x' for the ions formed?

Because one formula unit of the salt breaks into 'x' individual ions. So, if $0.1 \times 0.9$ moles of the salt dissociate, they will produce $x$ times that amount of ions.

Q: Is it necessary to convert the mass of water to grams?

Yes! The molar mass of water is $18 \text{ g/mol}$. To find the number of moles, the mass must be in the same units as the molar mass. So, $1.8 \text{ kg}$ becomes $1800 \text{ g}$.

Q: Can we approximate $0.01 + 0.09x$ to just $0.09x$?

In this specific case, no. The value of $0.01$ is comparable to $0.09x$ (since $x$ turns out to be 5, $0.09(5) = 0.45$). Ignoring it would lead to a significant error in the final integer answer.

Question

An aqueous solution is prepared by dissolving $0.1	ext{ mol}$ of an ionic salt in $1.8	ext{ kg}$ of water at $35^\circ	ext{C}$. The salt remains $90\%$ dissociated in the solution. The vapour pressure of the solution is $59.724	ext{ mm of Hg}$. Vapor pressure of water at $35^\circ	ext{C}$ is $60.000	ext{ mm of Hg}$. The number of ions present per formula unit of the ionic salt is ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are in a chemistry lab, holding a beaker filled with exactly $1.8 	ext{ kg}$ of pure water. At $35^\circ	ext{C}$, this pure water has a specific tendency to evaporate, creating a vapour pressure of $60.000 	ext{ mm Hg}$.

Now, you drop $0.1 	ext{ mol}$ of a mysterious ionic salt into the beaker. As it dissolves, it disrupts the water molecules' ability to escape into the vapour phase.

Consequently, the vapour pressure drops to $59.724 	ext{ mm Hg}$. Our mission is to figure out exactly how many ions each formula unit of this salt breaks into.

### Decoding the Dissociation

Before we dive into the vapour pressure math, we need to understand what's happening inside the solution. The salt doesn't just sit there; it dissociates. But there is a catch—it only dissociates $90\%$.

Let's assume one formula unit of the salt produces $x$ ions. We can represent this chemical process as:

$A_x \longrightarrow x A$

Initially, we have $0.1 	ext{ mol}$ of the salt. Since the degree of dissociation $\alpha$ is $0.9$, the amount of salt that actually breaks apart is $0.1 	imes 0.9 = 0.09 	ext{ mol}$.

This leaves behind $0.01 	ext{ mol}$ of undissociated salt. The part that did break apart forms $0.09x 	ext{ mol}$ of ions.

Therefore, the **total moles of non-volatile solute particles** floating in the water is the sum of the undissociated salt and the newly formed ions:

$n_{	ext{total}} = 0.01 + 0.09x$

### The Master Equation: Relative Lowering of Vapour Pressure

Now, let's connect the microscopic particles to the macroscopic vapour pressure. According to Raoult's Law, the relative lowering of vapour pressure is equal to the mole fraction of the solute.

However, to make our calculations significantly faster, we use a mathematically equivalent, modified form of the equation:

$\frac{P^\circ - P_s}{P_s} = \frac{n_{	ext{total}}}{N_{	ext{water}}}$

This form is brilliant because it removes the pesky addition term in the denominator that usually complicates the algebra.

We already know $n_{	ext{total}}$. What about $N_{	ext{water}}$? We have $1.8 	ext{ kg}$ of water, which is $1800 	ext{ g}$. Dividing by the molar mass of water ($18 	ext{ g/mol}$), we get exactly $100 	ext{ mol}$ of water.

### The Final Calculation

Let's plug all our known values into the master equation:

$\frac{60.000 - 59.724}{59.724} = \frac{0.01 + 0.09x}{100}$

The numerator simplifies to $0.276$. Now, we cross-multiply to solve for $x$:

$0.276 	imes 100 = 59.724 	imes (0.01 + 0.09x)$

$27.6 = 0.59274 + 5.33466x$

Subtracting $0.59274$ from both sides gives:

$27.00726 = 5.33466x$

Dividing to isolate $x$, we find that $x \approx 5.06$. Since the number of ions must be an integer, we can confidently round this to $5$.

**The salt produces 5 ions per formula unit.** This means our mysterious salt could be something like $Al_2(SO_4)_3$, which breaks into two Aluminum ions and three Sulfate ions!

Enter Numerical Value:

Visualized Solution

$Given Data$

$Moles of Solvent$

$Dissociation of Salt$

$Total Moles of Solute Particles$

$Relative Lowering of Vapour Pressure$

$Solving for x$

$Final Answer$

$The Way Forward$

The Sigma Insight: Colligative Properties

Analyzing the Setup

Decoding the Dissociation

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; } Given Data

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relative Lowering of Vapour Pressure

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

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

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