At \(0.00{ }^{\circ} \mathrm{C}\) and sea level, dry air can be assumed to have a molar volume of \(22.414 \mathrm{~L}\). Given that air is \(21.0 \%\) by volume oxygen, what is the molar concentration of oxygen in the atmosphere at sea level at \(0.00^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The molar concentration of oxygen in the atmosphere at sea level at 0.00°C is \(0.21 \mathrm{~mol/L}\).

Step by step solution

01

Calculate the molar volume of oxygen in air

We are given that air is 21.0% by volume oxygen. To find the molar volume of oxygen in air, we can simply multiply the molar volume of air by the percentage of oxygen. Molar volume of oxygen in air = (Molar volume of air) × (Percentage of oxygen) \(= 22.414 \mathrm{~L} \times 21.0 \% \)
02

Convert the percentage to decimal

Percentage should be converted to decimal before the multiplication: 21.0% = 0.21
03

Multiply the molar volume of air by the decimal percentage of oxygen

Now, multiply the molar volume of air by the decimal percentage of oxygen: Molar volume of oxygen in air = 22.414 L × 0.21 \(= 4.70694 \mathrm{~L} \)
04

Calculate the molar concentration of oxygen

Since we know the molar volume of oxygen in air, we can find the molar concentration of oxygen. Molar concentration is defined as the number of moles per unit volume: Molar Concentration of Oxygen = (Number of moles of Oxygen)/(Volume of Oxygen in L) We need to consider that 1 mole of any ideal gas occupies a volume of 22.414 L at 0.00°C and sea level. Therefore, the number of moles of oxygen will be: Number of moles of Oxygen = (Volume of Oxygen in L)/(Molar Volume of 1 mole of Oxygen) \( = 4.70694 \mathrm{~L} / 22.414 \mathrm{~L} \) Molar Concentration of Oxygen = (Number of moles of Oxygen)/(Volume of Oxygen in L) \( = \frac{4.70694 \mathrm{~L} / 22.414 \mathrm{~L}}{4.70694 \mathrm{~L}} \)
05

Simplify the expression

Now, simplify the expression by dividing: Molar Concentration of Oxygen = 0.21 mol/L So, the molar concentration of oxygen in the atmosphere at sea level at 0.00°C is 0.21 mol/L.

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