Problem 128 - A Complete Resource Book in Chemistry for JEE Main

Question

The amount of dissolved oxygen in 1 litre water in equilibrium with air at 1 atm pressure at \(25^{\circ} \mathrm{C}\) will be (assume that air contains 20 mole \% oxygen, Henry's constant \((\mathrm{kH})\) for oxygen is \(3.04 \times 10^{7} \mathrm{~mm} \mathrm{Hg}\) and density of \(\mathrm{H}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C}\) is \(1 \mathrm{~g} / \mathrm{cc}\) ) (a) the mole fraction of \(\mathrm{O}_{2}\) in solution is \(5 \times 10^{-6}\) (b) the mole fraction of \(\mathrm{O}_{2}\) in solution is \(5 \times 10^{-7}\) (c) the molarity of \(\mathrm{O}_{2}\) in solution is \(2.77 \times 10^{-5} \mathrm{M}\) (d) the molarity of \(\mathrm{O}_{2}\) in solution is \(2.77 \times 10^{-4} \mathrm{M}\)

Step-by-Step Solution

Answer

The mole fraction of \(O_2\) is \(5 \times 10^{-6}\) and the molarity is \(2.77 \times 10^{-4} \text{ M}\). Answer: options (a) and (d).

Step 1: Calculate Partial Pressure of Oxygen

Given that air contains 20 mole % oxygen. To find the partial pressure of oxygen in air at 1 atm, use the equation: \[ P_{O_2} = 0.20 \times 1 \space ext{atm} = 0.20 \space ext{atm} \] Convert atm to mm Hg, knowing that \(1 \text{ atm} = 760 \text{ mm Hg}\), thus: \[ P_{O_2} = 0.20 \times 760 = 152 \text{ mm Hg} \]

Step 2: Apply Henry's Law to Find the Mole Fraction

Henry's Law states \( C = k_H \times P \), where \( C \) is the concentration of the gas (in molarity), \( k_H \) is Henry's constant, and \( P \) is the partial pressure.Using Henry's constant \( k_H = 3.04 \times 10^7 \text{ mm Hg} \):\[ x_{O_2} = \frac{P_{O_2}}{k_H} = \frac{152}{3.04 \times 10^7} \approx 5 \times 10^{-6} \]

Step 3: Relate Mole Fraction to Molarity

To find molarity, use the formula: \[ \text{Molarity, M} = x_{O_2} \times \frac{d}{M_{H_2O}} \times 1000 \] where \(d\) is the density of water (1 g/cc or 1000 g/L) and \(M_{H_2O}\) is the molar mass of water (approximately 18 g/mol).Plugging in values: \[ M = 5 \times 10^{-6} \times \frac{1000}{18} \approx 2.77 \times 10^{-4} \text{ M} \]

Step 4: Compare Solutions to Answer Choices

Based on the calculations, we have the following:- Mole fraction of \(O_2\) in the solution is \(5 \times 10^{-6}\) which matches option (a).- Molarity of \(O_2\) in the solution is \(2.77 \times 10^{-4} \text{ M}\) which corresponds to option (d).

Key Concepts

Partial Pressure

Partial pressure is a key concept in understanding how gases behave in mixtures like air. When you're dealing with gases in a mixture, rather than thinking of their total pressure, it's sometimes easier to consider each gas's pressure separately. That's the essence of partial pressure: it's the pressure a gas would exert if it were the only one there in a given volume.

Partial pressure is calculated based on the percentage of the gas present in the mixture. For instance, if oxygen makes up 20% of the air, like in our problem, its partial pressure would be 20% of the total atmospheric pressure. Normally, at sea level, atmospheric pressure is 1 atmosphere (atm), and since 1 atm equals 760 millimeters of mercury (mm Hg), you find the partial pressure of oxygen by this calculation:

• Multiply the percentage (0.20 for oxygen) by the total pressure in atm: \[ P_{O_2} = 0.20 \times 1 \text{ atm} = 0.20 \text{ atm} \]
• Convert this to mm Hg: \[ 0.20 \times 760 = 152 \text{ mm Hg} \]

This tells us that the pressure from the oxygen alone in the air is 152 mm Hg.

Mole Fraction

The mole fraction is another important concept in chemical solutions and understanding Henry's Law. It represents the ratio of the number of moles of a particular component to the total number of moles of all components in the mixture. Mole fraction is dimensionless, meaning it has no units.

In the context of gases dissolved in a liquid, like oxygen in water, Henry's Law helps us connect partial pressure with mole fraction. According to Henry's Law, the concentration of a gas in a liquid (which can relate to mole fraction or molarity) is proportional to the partial pressure of the gas above the liquid. Mathematically, it's expressed as:

• Henry's Law: \[ C = k_H \times P \] where \( C \) is concentration, usually in molarity, \( k_H \) is Henry's Law constant, and \( P \) is the partial pressure.
• Rearranging to find mole fraction: \[ x_{O_2} = \frac{P_{O_2}}{k_H} \]

Applying the values from the problem, we find:

• Mole fraction of oxygen in water: \[ x_{O_2} = \frac{152}{3.04 \times 10^7} \approx 5 \times 10^{-6} \]

This shows how little oxygen actually dissolves in water at a time.

Molarity Calculation

Molarity is a common way to express the concentration of a solution, representing the number of moles of solute per liter of solution. Understanding how to convert mole fraction to molarity is crucial when applying Henry's Law for practical purposes.

To derive molarity from the mole fraction using the known density of water, we need to bridge the gap between these concepts.

• Use this formula: \[ \text{Molarity}, M = x_{O_2} \times \frac{d}{M_{H_2O}} \times 1000 \]
• Where:
  • \( x_{O_2} \) is the mole fraction.
  • \( d \) is the density of water, 1 g/cc or 1000 g/L.
  • \( M_{H_2O} \) is the molar mass of water, approximately 18 g/mol.
• Insert values to calculate: \[ M = 5 \times 10^{-6} \times \frac{1000}{18} \approx 2.77 \times 10^{-4} \text{ M} \]

This calculation tells us that the molarity of dissolved oxygen is very small, indicating low solubility of oxygen at room temperature and pressure.