A gas diffuses only 0.5 times as fast as an oxygen molecule at the same temperature. What is the molecular weight of the gas?
(rate1/rate2)=sqrt(MW2/MW1).
I would make up a rate for oxygen and use half of that rate for the unknown gas. Post your work if you get stuck.
(2.0/0.5)=4
im confused i cant set this problem up correctly? I need a lot of help
Meaning you want me to do all the thinking?
rate 1 = rate oxygen = 2 liters/hour.
rate 2 = rate unknown = (1/2)*2 = 1 liter/hour.
Mol Weight 1 = mw oxygen = 32
mol wt 2 = mw unknown.
(rate1/rate2) = sqrt (M2/M1).
Substitute and solve for M2
if the equation is right:
2/1=square root x/32
so square both sides
4=x/32
x=4*32
x=128 or tellurium
Then your answer is B
To determine the molecular weight of the gas, we need to understand Graham's law of diffusion. Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
The formula for Graham's law of diffusion is:
Rate1 / Rate2 = √(Molar Mass2 / Molar Mass1)
In this case, we are given that the gas diffuses only 0.5 times as fast as an oxygen molecule. Let's assign the following variables:
Rate1 = rate of diffusion for the oxygen molecule
Rate2 = rate of diffusion for the gas
Molar Mass1 = molar mass of the oxygen molecule
Molar Mass2 = molar mass of the gas (which we need to find)
Using the given information, we can set up the equation as follows:
0.5 = √(Molar Mass2 / Molar Mass1)
To solve for the molecular weight of the gas, we need to isolate Molar Mass2. We can square both sides of the equation:
(0.5)^2 = (Molar Mass2 / Molar Mass1)
0.25 = (Molar Mass2 / Molar Mass1)
Next, we can cross-multiply and solve for Molar Mass2:
0.25 * Molar Mass1 = Molar Mass2
Therefore, the molecular weight of the gas is 0.25 times the molecular weight of the oxygen molecule.