You have 4.271 x 102 mL of oxygen gas (O2) at 7.656 x 102 mm of Hg and 78 oC. What is the mass of the sample in milligrams?
What is the ratio of effusion rates for a light gas, such as HF, to a heavy gas, such as UF4?
To calculate the mass of the sample, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature of the gas (in Kelvin)
First, we need to convert the given pressure from mmHg to atm and the given temperature from degrees Celsius to Kelvin.
Given:
Pressure (P) = 7.656 x 10^2 mmHg
Temperature (T) = 78 °C
To convert mmHg to atm, divide the pressure by 760:
P = 7.656 x 10^2 mmHg / 760 mmHg/atm
P = 1.006 atm
To convert Celsius to Kelvin, add 273.15 to the temperature:
T = (78 + 273.15) K
T = 351.15 K
Now, we need to rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Plugging in the values:
n = (1.006 atm) * (4.271 x 10^2 mL / 1000 mL/L) / (0.0821 L.atm/mol.K) * 351.15 K
Convert mL to L by dividing by 1000:
n = (1.006 atm) * (4.271 x 10^2 mL / 1000 mL/L) / (0.0821 L.atm/mol.K) * 351.15 K
n = (1.006 atm) * (0.4271 L) / (0.0821 L.atm/mol.K) * 351.15 K
n ≈ 0.02083 moles
Now, to calculate the mass of the sample in milligrams, we need to use the molar mass of oxygen gas (O2). The molar mass of O2 is approximately 32 g/mol.
To convert moles to grams, multiply the number of moles by the molar mass:
mass = n * molar mass
mass = 0.02083 moles * 32 g/mol
mass = 0.66656 g
Finally, to convert grams to milligrams, multiply the mass by 1000:
mass = 0.66656 g * 1000 mg/g
mass ≈ 666.56 mg
Therefore, the mass of the sample is approximately 666.56 milligrams.
Now, let's move on to the second question.
The ratio of effusion rates of two gases can be calculated using Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
The ratio can be calculated as follows:
Rate of effusion (R1) / Rate of effusion (R2) = √(Molar mass (M2) / Molar mass (M1))
Given that HF is the light gas and UF4 is the heavy gas, let's represent them as follows:
R1 = effusion rate of HF
R2 = effusion rate of UF4
M1 = molar mass of HF
M2 = molar mass of UF4
To find the ratio, we need to know the molar masses of HF and UF4.
The molar mass of HF can be calculated by adding the molar masses of hydrogen (H) and fluorine (F):
M1 = Molar mass of H + Molar mass of F
M1 = (1.008 g/mol) + (18.998 g/mol) ≈ 20.006 g/mol
The molar mass of UF4 can be calculated by adding the molar masses of uranium (U) and four fluorine (F) atoms:
M2 = Molar mass of U + 4 * Molar mass of F
M2 = (238.029 g/mol) + 4 * (18.998 g/mol) ≈ 352.025 g/mol
Now, we can calculate the ratio of effusion rates:
Ratio = √(M2 / M1)
Ratio = √(352.025 g/mol / 20.006 g/mol)
Ratio ≈ √(17.596)
Therefore, the ratio of effusion rates for a light gas (HF) to a heavy gas (UF4) is approximately √(17.596).