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 He(g) is present in a helium-oxygen mixture having a density of 0.538g/L at 25celsius and 721mmHg ?
change that density to stp.
density1=P2T2/(P1*T1 density2)
Then
at stp, using decimalpercents,
densityHe*percentHe+densityO2*percentO2 =denstiygasSTP
Those percents will be by mass. To change to percent by moles, divide each percent by mass by atomic mass of He, and O2, and normalize to 100.
wat do u mean by change density to stp
To determine the percent (by moles) of He(g) present in the helium-oxygen mixture, we can use the ideal gas law and the concept of partial pressure.
First, let's convert the given density of the mixture to its molar mass. We can assume that the mixture is at STP (Standard Temperature and Pressure), where the temperature is 0 degrees Celsius and the pressure is 1 atm.
The molar mass (M) of the mixture can be calculated using the formula:
M = density * (RT/P)
Where:
- R is the ideal gas constant (0.0821 L·atm/mol·K)
- T is the temperature in Kelvin
- P is the pressure in atm
Converting the given temperature of 25 degrees Celsius to Kelvin:
T = 25 + 273.15 = 298.15 K
Now, calculate the molar mass:
M = 0.538 g/L * (0.0821 L·atm/mol·K * 298.15 K)/(721 mmHg * (1 atm/760 mmHg))
Simplifying the equation:
M = 0.538 g/L * 0.08498 mol/L = 0.0458 mol/L
Next, we need to find the partial pressure of He(g) in the mixture. The fraction of moles of He(g) in the mixture (x) can be determined using Dalton's law of partial pressures:
P_He = x * P_total
Where:
- P_He is the partial pressure of He(g)
- P_total is the total pressure of the mixture
We are given that the total pressure is 721 mmHg. To find the partial pressure of He(g), we need to know the mole fraction of He(g) in the mixture.
x = n_He / (n_He + n_O2)
Where:
- n_He is the number of moles of He(g)
- n_O2 is the number of moles of O2(g)
We don't have the number of moles of He(g) or O2(g) explicitly given. However, we can use the molar mass of the mixture (0.0458 mol/L) to calculate the moles of He(g) and O2(g) per liter of mixture.
Let's assume that the molar mass of O2(g) is approximately 32 g/mol. Since the molar mass of the mixture is due to both helium and oxygen, the rest must be attributed to oxygen:
n_O2 = (M - m_He) / M * mol/L
Substituting the given values:
n_O2 = (0.0458 mol/L - m_He) / 0.0458 mol/L
Now, substituting this into the equation for mole fraction:
x = n_He / (n_He + n_O2) = n_He / (n_He + (0.0458 mol/L - m_He) / 0.0458 mol/L)
Simplifying the equation:
x = n_He / (0.0458 mol/L + (0.0458 mol/L - m_He))
Since the mole fraction (x) is defined as the ratio of moles of component A to the total moles of all components, we can write:
x = n_He / (n_He + (0.0458 mol/L - n_He)) = n_He / 0.0458 mol/L
Simplifying the equation:
x = n_He / 0.0458 mol/L = (n_He * 0.0458 mol/L) / (n_He) = 0.0458 mol/mol
Therefore, the percent (by moles) of He(g) present in the helium-oxygen mixture is approximately 4.58%.