I'm stumped again :(
the question is - A gas sprayed from an aerosol can effuses 5.06 times slower than oxygen.If the molar mass of oxygen is 32.g/mol, what is the molar mass of the unknown gas?
I don't know if I am doing the ideal gas law or grahams law or how to set it up. Please help. Thanks
Graham's Law is what you want.
rate 1/rate 2 = (sqrt molar mass 2/sqrt molar mass 1)
so what would be rate 1 and 2?
Make up a number for the rate of the unknown gas. Then oxygen will be 5.06 times less than that.
No worries, I'm here to help you out!
To solve this problem, you need to use Graham's law of effusion, which relates the rates of effusion of two gases to their molar masses. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Here's how you can set up the equation using Graham's law:
Let's assume the molar mass of the unknown gas is "x" g/mol.
According to the problem, the gas effuses 5.06 times slower than oxygen. So, the rate of effusion of the unknown gas, divided by the rate of effusion of oxygen, is equal to 1/5.06:
(rate of effusion of unknown gas) / (rate of effusion of oxygen) = 1/5.06
Next, you need to use Graham's law, which states that the rate of effusion of a gas is proportional to the inverse square root of its molar mass. Thus, the rate of effusion of the unknown gas divided by the rate of effusion of oxygen is equal to the square root of the ratio of their molar masses:
(rate of effusion of unknown gas) / (rate of effusion of oxygen) = √(molar mass of oxygen / molar mass of unknown gas)
Now, substitute the known values into the equation. The molar mass of oxygen is given as 32 g/mol:
(rate of effusion of unknown gas) / (rate of effusion of oxygen) = √(32 g/mol / x g/mol)
Simplify the right side of the equation:
(rate of effusion of unknown gas) / (rate of effusion of oxygen) = √(32 / x)
Now, plug in the given ratio of 1/5.06:
1/5.06 = √(32 / x)
To solve for x, square both sides of the equation:
1/(5.06)^2 = 32 / x
Simplify and solve for x:
x = 32 / (1/(5.06)^2)
Calculating this expression will give you the molar mass of the unknown gas in grams per mole.
I hope this helps you set up the equation correctly and solve the problem!