Dry air (0% humidity) is approximately 78% N2, 21% O2, and 1% Ar by moles. Calculate the molar mass and density of dry air at 0 °C and 1.0 atm. Assume ideal behavior and use two significant figures in your answers.
I calculated the molar mass to be approximately 29 g/mol. Although I am not entirely sure whether that is the correct answer or not.
29 is about right but I can't say you did it right since you didn't show your work.
The density = 29/22.4 = ? g/L
28.02g N2
32.00 g O2
39.95 g Ar
Using the percentage abundances of each gas, multiply each gases abundance by their molar mass.
molar mass of air=28.02(.78)+32(.21)+39.95(.01) = 28.98 g/mol.
using two sig figs =29 g/mol
dry air 0% humidity is approximately 78% N2, 21% 02,1% Ar by moles. calculate thew molar massand density of dry air at 0% at 1atm?
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To calculate the molar mass and density of dry air at 0 °C and 1.0 atm, we first need to find the molar mass of each component (N2, O2, and Ar) and then calculate the average molar mass of dry air.
1. Molar Mass of N2:
The molar mass of N2 is 28.02 g/mol (14.01 g/mol for each nitrogen atom).
2. Molar Mass of O2:
The molar mass of O2 is 32.00 g/mol (16.00 g/mol for each oxygen atom).
3. Molar Mass of Ar:
The molar mass of Ar is 39.95 g/mol.
To calculate the average molar mass of dry air, we need to consider the mole fractions of each component. Given that dry air is approximately 78% N2, 21% O2, and 1% Ar, we can calculate the mole fractions:
- Mole Fraction of N2 (X_N2) = 0.78 (78% in decimal form)
- Mole Fraction of O2 (X_O2) = 0.21 (21% in decimal form)
- Mole Fraction of Ar (X_Ar) = 0.01 (1% in decimal form)
Next, we can calculate the average molar mass (M_avg) using the formula:
M_avg = X_N2 * M_N2 + X_O2 * M_O2 + X_Ar * M_Ar
Plugging in the values:
M_avg = (0.78 * 28.02 g/mol) + (0.21 * 32.00 g/mol) + (0.01 * 39.95 g/mol)
M_avg ≈ 28.97 g/mol
So, the molar mass of dry air at 0 °C and 1.0 atm is approximately 28.97 g/mol.
To calculate the density (ρ) of dry air, we can use the Ideal Gas Law:
PV = nRT
Where:
P = Pressure (1.0 atm)
V = Volume (assume 1 mole of air)
n = Number of moles of air (1.0 mole)
R = Ideal Gas Constant (0.0821 L.atm/mol.K)
T = Temperature (0 °C + 273.15) = 273.15 K
Rearranging the formula:
ρ = (n * M_avg) / V
Substituting the known values:
ρ = (1.0 mol * 28.97 g/mol) / 22.414 L*mol⁻¹
Calculating:
ρ ≈ 1.29 g/L
So, the density of dry air at 0 °C and 1.0 atm is approximately 1.29 g/L, with two significant figures.