At 273 K ,23.54mL of N₂ at a partial pressure of 1atm will dissolve per litre of water.Under the same conditions,48.89mL of O₂ will dissolve.Calculate the amount by which the freezing point of air free water is lowered by being saturated with air.(composition of air is 79 vol % of N₂ and 21 vol % of O₂)K_f=1.86K kg mol^-1.

To determine how much the freezing point of water is lowered when it is saturated with air, we first need to calculate the molality of the dissolved gases (N2 and O2) in the water. This involves a few steps, including finding the number of moles of each gas that dissolve in a liter of water and then using the freezing point depression formula.

Step 1: Calculate Moles of Dissolved Gases

We know that at 273 K (0 °C), the solubility of N2 is 23.54 mL per liter of water, and for O2, it is 48.89 mL per liter. To convert these volumes into moles, we can use the ideal gas law, which states that at standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L.

Calculating Moles of N2

• Volume of N2 = 23.54 mL = 0.02354 L
• Moles of N2 = Volume / Molar Volume = 0.02354 L / 22.4 L/mol ≈ 0.00105 mol

Calculating Moles of O2

• Volume of O2 = 48.89 mL = 0.04889 L
• Moles of O2 = Volume / Molar Volume = 0.04889 L / 22.4 L/mol ≈ 0.00218 mol

Step 2: Total Moles of Gases

Now, we can find the total number of moles of gases dissolved in 1 liter of water:

• Total moles = Moles of N2 + Moles of O2 = 0.00105 mol + 0.00218 mol ≈ 0.00323 mol

Step 3: Calculate Molality

Molality (m) is defined as the number of moles of solute per kilogram of solvent. Since we are considering 1 liter of water, which has a mass of approximately 1 kg, the molality is:

• Molality = Total moles of gases / Mass of water (kg) = 0.00323 mol / 1 kg = 0.00323 m

Step 4: Freezing Point Depression Calculation

The freezing point depression can be calculated using the formula:

ΔTf = Kf × m

Where:

• ΔTf = change in freezing point
• Kf = freezing point depression constant (1.86 K kg mol^-1)
• m = molality (0.00323 m)

Now, substituting the values:

• ΔTf = 1.86 K kg mol^-1 × 0.00323 m ≈ 0.00601 K

Final Result

The freezing point of water is lowered by approximately 0.00601 K when saturated with air. This means that the freezing point of the water would be around 0 °C - 0.00601 K, which is slightly below the normal freezing point of pure water.

This calculation illustrates how the presence of dissolved gases can affect the physical properties of water, such as its freezing point. In practical terms, this is a small effect, but it is significant in various scientific and environmental contexts.