Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 5.36 mL of O2 had passed through the membrane, but only 3.60 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas?
Let's assume the time is 1 sec. Then rate O2 = 5.36 mL/sec.
rate unknown gas = 3.60 mL/sec
rate O2/rate unk) = sqrt(mm unk/mmO2)
Solve for mm unk.
To find the molar mass of the unknown gas, we can use Graham's Law of Effusion. Graham's Law states that the rate of effusion (the amount of gas that passes through a membrane in a given time) is inversely proportional to the square root of the molar mass.
First, we need to find the ratio of the rates of effusion between the two gases.
Let's call the molar mass of the unknown gas "Mx".
According to Graham's Law,
Rate of effusion of O2 / Rate of effusion of unknown gas = Square root of (Molar mass of unknown gas / Molar mass of O2)
Since the rates of effusion are given as volumes, we can rewrite the equation as:
Volume of O2 / Volume of unknown gas = Square root of (Molar mass of unknown gas / Molar mass of O2)
Plugging in the given values:
5.36 mL / 3.60 mL = √(Mx / 32.00 g/mol)
Now, let's solve for the molar mass of the unknown gas (Mx):
Cross-multiply the equation:
5.36 mL * 32.00 g/mol = 3.60 mL * Mx
171.52 g * mL/mol = 3.60 mL * Mx
Divide both sides by 3.60 mL to isolate Mx:
Mx = (171.52 g * mL/mol) / 3.60 mL
Mx = 47.64 g/mol
Therefore, the molar mass of the unknown gas is 47.64 g/mol.