Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of Helium is present in a helium-oxygen mixture having a density of 0.538 g/L at 25 degrees C and 721 mmHg?
Express your answer numerically as a percentage.
sorry. didn't mean to post twice
To find the percent by moles of helium in the mixture, we need to calculate the mole fraction of helium.
First, let's calculate the mole fraction of helium using the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm (721 mmHg = 721/760 atm)
V = volume in liters (0.538 L)
n = moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin (25 + 273 = 298 K)
Rearranging the equation to solve for n, we have:
n = PV / RT
n = (721/760) * (0.538) / (0.0821 * 298)
n ≈ 0.2008 moles
Next, let's calculate the mole fraction of helium using the moles of helium and oxygen present in the mixture.
Given:
Mole fraction of helium (XHe) + Mole fraction of oxygen (XO2) = 1
Let's assume the mole fraction of helium is XHe and the mole fraction of oxygen is XO2.
Since helium and oxygen are the only two components in the mixture, we can express their mole fractions as follows:
XHe = moles of helium / total moles
XO2 = moles of oxygen / total moles
Plugging in the values we calculated earlier:
XHe = 0.2008 / (0.2008 + 1)
XHe = 0.2008 / 1.2008 ≈ 0.1672
Finally, let's express the mole fraction of helium as a percentage:
Percentage = XHe * 100
Percentage ≈ 0.1672 * 100 ≈ 16.72%
Therefore, the percent (by moles) of helium present in the helium-oxygen mixture is approximately 16.72%.
To find the percent (by moles) of Helium in the helium-oxygen mixture, we need to use the Ideal Gas Law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, we need to determine the number of moles of the mixture. We can use the formula:
n = PV / RT
Given:
Pressure (P) = 721 mmHg
Volume (V) = 1 L (since the question gives the density in g/L)
Temperature (T) = 25 degrees C = 25 + 273.15 = 298.15 K
Ideal Gas Constant (R) = 0.0821 L · atm/K · mol
n = (721 mmHg * 1 L) / (0.0821 L · atm/K · mol * 298.15 K)
Now, let's calculate the moles of the mixture.
n = 29.4 mol
Next, we need to find the moles of Helium in the mixture. Let's assume the mole fractions of helium and oxygen are x and y, respectively. Since the mixture consists of only helium and oxygen, x + y = 1.
The mole fraction of helium can be written as:
x = n(He) / n(total)
The density of the mixture is given as 0.538 g/L, but we can convert it into grams per mole by using the molar mass of the mixture. Helium has a molar mass of 4.0026 g/mol, and oxygen has a molar mass of 31.9988 g/mol.
Molar mass of the mixture = x(He) * molar mass(He) + y(O2) * molar mass(O2)
0.538 g/L = x * 4.0026 g/mol + (1 - x) * 31.9988 g/mol
Simplifying the equation:
0.538 g/L = 4.0026x g/mol + (31.9988 - 31.9988x) g/mol
0.538 g/L = 31.9988 - 27.9962x
27.9962x = 31.9988 - 0.538
x = (31.9988 - 0.538) / 27.9962
Now, let's calculate the value of x, which represents the mole fraction of helium.
x = 0.11052
As the mole fraction represents the ratio of moles, we can convert it to a percentage by multiplying by 100.
Percentage of Helium = x * 100
Now, let's calculate the percentage of Helium in the mixture.
Percentage of Helium = 0.11052 * 100 = 11.05%
Therefore, the percentage (by moles) of Helium in the helium-oxygen mixture is approximately 11.05%.