Two open beakers one containing a solvent and the other containing a mixture of that solvent with a

Q: Why does the solute lower the vapour pressure?

Non-volatile solute particles occupy space at the surface of the liquid, reducing the number of solvent molecules that can escape into the vapour phase.

Q: Will the transfer of vapour ever stop?

Yes, it will stop when the vapour pressure of the solution becomes equal to the vapour pressure of the pure solvent. However, since the pure solvent's vapour pressure remains constant at a given temperature, the pure solvent will completely evaporate into the solution if the container is small enough!

Q: What if the solute was volatile?

If the solute was volatile, it would also contribute to the total vapour pressure. The net transfer would depend on the relative volatilities and mole fractions of both components.

Q: Does the temperature of the system change during this process?

Evaporation is an endothermic process (absorbs heat), and condensation is exothermic (releases heat). In a perfectly insulated container, the pure solvent would cool down slightly, and the solution would warm up slightly, but we usually assume the container is in thermal equilibrium with its surroundings.

Q: Is this related to osmosis?

Absolutely! In osmosis, solvent flows from a region of higher chemical potential (pure solvent) to lower chemical potential (solution) through a semi-permeable membrane. Here, the 'membrane' is the vapour space, and the driving force is the difference in vapour pressure!

Question

Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non-volatile solute are together sealed in a container. Over time

Options:

(A) the volume of the solution decreases and the volume of the solvent increases

(B) the volume of the solution does not change and the volume of the solvent decreases

(C) the volume of the solution increases and the volume of the solvent decreases

(D) the volume of the solution and the solvent does not change

Sigma Percentile · Accepted Answer

### The Thought Experiment

Imagine a sealed glass container. Inside, we place two open beakers side by side. One beaker contains a pure solvent, like water. The other beaker contains a solution—the same solvent, but with a non-volatile solute dissolved in it, like salt or sugar.

What happens over time? To the naked eye, it might seem like nothing. But at the microscopic level, a fierce battle of thermodynamics is taking place!

### Raoult's Law and Vapour Pressure

To understand the invisible forces at play, we must look at the surface of the liquids. In both beakers, solvent molecules are constantly escaping into the air (evaporation) and returning to the liquid (condensation).

For the pure solvent, this establishes an equilibrium vapour pressure, which we call $p^\circ$.

However, in the second beaker, the non-volatile solute particles occupy precious real estate at the surface. This physical blockade means fewer solvent molecules can escape. According to **Raoult's Law**, the vapour pressure of the solution ($p_s$) is directly proportional to the mole fraction of the solvent ($x_A$).

$$p_s = p^\circ \cdot x_A$$

Since the mole fraction of the solvent in a solution is always less than 1 ($x_A < 1$), the vapour pressure of the solution is strictly less than that of the pure solvent ($p_s < p^\circ$).

### The Invisible Transfer

Now, here is where the magic happens. Both beakers are sharing the same sealed airspace. The pure solvent is trying to fill the space with enough vapour to reach a pressure of $p^\circ$.

Meanwhile, the solution is trying to maintain a lower pressure of $p_s$.

Because $p^\circ > p_s$, the actual pressure in the container will settle somewhere in between. This creates a fascinating dynamic:

1. The pressure is **too low** for the pure solvent, so it evaporates faster than it condenses.
2. The pressure is **too high** for the solution, so it condenses vapour faster than it evaporates.

### The Final Outcome

The result is a continuous, invisible river of vapour flowing from the pure solvent to the solution.

Over time, the pure solvent loses mass, causing its volume to **decrease**. The solution, acting like a sponge for the vapour, gains mass, causing its volume to **increase**.

This beautiful phenomenon is essentially **isothermal distillation**. It will continue until the vapour pressures equalize—which, in this case, means the pure solvent will completely evaporate, leaving its beaker bone dry!

Animated Solution for Chemistry - Solutions: Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non-volatile solute are together sealed in a container. Over time

Select Answer:

Visualized Solution

$System Setup$

$p_{s} = p^{\circ} x_{A}$

$p^{\circ} > p_{s}$

$Δ m \neq = 0$

$Δ V_{1} < 0, Δ V_{2} > 0$

$p_{f ina l} = ?$

The Sigma Insight: Henry's Law and Raoult's Law

The Thought Experiment

Raoult's Law and Vapour Pressure

The Invisible Transfer

The Final Outcome

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