The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The mol

Q: Why did we equate mole fraction to molar solubility?

Strictly speaking, they are different. Molarity is moles per liter, while mole fraction is dimensionless. However, in this specific JEE question, the numerical value of the mole fraction was intended as the answer. Always be prepared to adapt if strict dimensional analysis doesn't align with the given options or format.

Q: What if I wanted to calculate the actual molarity?

You would use the density of water ($1 \text{ kg/L}$) to find the moles of water in 1 Liter ($55.5 \text{ mol}$). Then, using the mole fraction, you'd find the moles of $O_2$, which would be approximately $0.0138 \text{ M}$.

Q: Does temperature affect Henry's constant?

Yes! $K_H$ is highly temperature-dependent. As temperature increases, $K_H$ typically increases, which means solubility decreases. That's why warm soda goes flat faster!

Q: Why is the partial pressure given in kPa instead of atm?

Pressure can be measured in various units. As long as the partial pressure and Henry's constant are in the same units (like kPa here), they will cancel out perfectly when calculating the dimensionless mole fraction.

Q: What does the density of water have to do with this problem?

In a perfectly rigorous calculation, you'd need the density to convert the mass of the solvent into volume to find molarity. Here, it's a distractor because the question's intended solution path bypasses the true molarity calculation.

Question

The oxygen dissolved in water exerts a partial pressure of $20 	ext{ kPa}$ in the vapour above water. The molar solubility of oxygen in water is ...... $	imes 10^{-5} 	ext{ mol dm}^{-3}$.
(Round off to the nearest integer).
[Given, Henry's law constant ($K_H$) $= 8.0 	imes 10^4 	ext{ kPa}$ for $	ext{O}_2$, density of water with dissolved oxygen $= 1.0 	ext{ kg dm}^{-3}$].

Sigma Percentile · Accepted Answer

### The Setup: Gas Meets Liquid

Imagine you are looking at a sealed bottle of water. Above the liquid surface, oxygen gas ($O_2$) is bouncing around, exerting a pressure. This pressure is known as the **partial pressure**, and in our problem, it is given as $20 	ext{ kPa}$.

Because of this pressure, some of the oxygen molecules are forced into the water, dissolving into the liquid. The relationship between the pressure of the gas above the liquid and the amount of gas that dissolves is governed by a beautiful and simple rule: **Henry's Law**.

### The Master Equation: Henry's Law

Henry's Law states that the partial pressure of a gas ($p$) is directly proportional to its mole fraction ($\chi$) in the solution. Mathematically, we write this as:

$$p = K_H \cdot \chi$$

Here, $K_H$ is the **Henry's Law constant**, which is specific to the gas, the solvent, and the temperature. The problem provides us with $K_H = 8.0 	imes 10^4 	ext{ kPa}$. Notice that both the partial pressure and the Henry's constant are in the same units (kPa). This is perfect because it means we don't need to do any messy unit conversions before plugging them into our equation.

### The Calculation: Finding the Mole Fraction

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

$$20 = (8.0 	imes 10^4) \cdot \chi_{O_2}$$

To isolate the mole fraction ($\chi_{O_2}$), we simply divide the pressure by the Henry's constant:

$$\chi_{O_2} = \frac{20}{8.0 	imes 10^4}$$

$$\chi_{O_2} = 2.5 	imes 10^{-4}$$

So, the mole fraction of oxygen in the water is $2.5 	imes 10^{-4}$.

### The Exam Trap: Mole Fraction vs. Molarity

Now, here is where we need to be smart about competitive exams. The question asks for the **molar solubility** in $	ext{mol dm}^{-3}$ (which is Molarity). Strictly speaking, mole fraction and molarity are not the same thing. Molarity is the number of moles of solute per liter of solution, while mole fraction is a dimensionless ratio of moles.

If we were to calculate the true molarity using the given density of water ($1.0 	ext{ kg dm}^{-3}$), we would find that 1 liter of water contains about $55.5 	ext{ moles}$. The true molarity would be approximately $0.0138 	ext{ M}$.

However, in the context of this specific JEE problem, the intended solution path equates the numerical value of the mole fraction directly to the requested solubility. This is a known dimensional inconsistency that occasionally appears in exams.

To match the requested format of $...... 	imes 10^{-5}$, we rewrite our mole fraction:

$$2.5 	imes 10^{-4} = 25 	imes 10^{-5}$$

Thus, the integer we need to fill in the blank is **25**.

Enter Numerical Value:

Visualized Solution

$S y s t e m S e t u p$

$H e n r y^{'} s L a w$

$S u b s t i t u t in g V a l u es$

$C a l c u l a t in g M o l e F r a c t i o n$

$M o l a r S o l u bi l i t y$

$D im e n s i o na l A na l y s i s N o t e$

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

The Setup

The Master Equation

The Calculation

The Exam Trap

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Comprehension Passage

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Enter Numerical Value:

Visualized Solution

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