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Chapter 11: Problem 87

What volume in milliliters of \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas at \(0.996 \mathrm{~atm}\) and \(25.0^{\circ} \mathrm{C}\) contains \(0.200 \mathrm{~g}\) of oxygen?

Short Answer

Expert verified
The volume of N₂O₄ gas containing 0.200 g of oxygen at 0.996 atm and 25.0°C is approximately 8.439 mL.

Step by step solution

01

Find the molar mass of N₂O₄

First, we need to find the molar mass of N₂O₄. To do this, we will add the molar masses of the individual elements in the molecule. The molecular formula for N₂O₄ is: - Nitrogen (N): 2 atoms with a molar mass of \( 14.01\,\mathrm{g/mol}\) - Oxygen (O): 4 atoms with a molar mass of \( 16.00\,\mathrm{g/mol}\) Molar mass of N₂O₄ = (2 × 14.01) + (4 × 16.00) = 92.02 g/mol
02

Calculate the moles of oxygen

Now that we have the molar mass of N₂O₄, we can find the moles of oxygen in 0.200 g. Moles of oxygen = \(\frac{\text{mass of oxygen}}{\text{molar mass of oxygen}}\) Moles of oxygen = \(\frac{0.200\,\mathrm{g}}{16.00\,\mathrm{g/mol}} = 0.0125\,\mathrm{mol} \)
03

Calculate the moles of N₂O₄

Since there are 4 moles of oxygen (O) in 1 mole of N₂O₄, we can now find out how many moles of N₂O₄ are required. Moles of N₂O₄ = \(\frac{\text{moles of oxygen}}{4}\) Moles of N₂O₄ = \(\frac{0.0125\,\mathrm{mol}}{4} = 0.003125\,\mathrm{mol} \)
04

Convert the temperature to Kelvin

To use the Ideal Gas Law, we need to convert the given temperature from Celsius to Kelvin. Temperature (K) = Temperature (°C) + 273.15 Temperature (K) = 25.0 + 273.15 = 298.15 K
05

Use the Ideal Gas Law to find the volume

Now that we have the number of moles of N₂O₄ and the temperature in Kelvin, we can use the Ideal Gas Law (PV = nRT) to find the volume. The ideal gas constant (R) is 0.0821 L atm/mol K. \(V = \frac{nRT}{P}\) \(V = \frac{0.003125\,\mathrm{mol} \times 0.0821\,\mathrm{L\cdot atm \cdot mol}^{-1}\cdot\mathrm{K}^{-1} \times 298.15\,\mathrm{K}}{0.996\,\mathrm{atm}}\) \(V = 0.008439\,\mathrm{L}\)
06

Convert the volume to milliliters

Finally, we'll convert the volume of N₂O₄ gas from liters to milliliters. Volume (mL) = Volume (L) × 1000 mL/L Volume (mL) = 0.008439 L × 1000 mL/L = 8.439 mL The volume of N₂O₄ gas containing 0.200 g of oxygen at 0.996 atm and 25.0°C is approximately 8.439 mL.

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