Final answer:
The molecular weight of a gas that diffuses 1/50 as fast as hydrogen can be calculated using Graham's law of effusion. By squaring the ratio of the rates of diffusion and multiplying by the molecular weight of hydrogen (2.016 g/mol), we find the molecular weight of the unknown gas to be approximately 5040 g/mol, which is unusually high for gas.
Step-by-step explanation:
To determine the molecular weight of a gas that diffuses 1/50 as fast as hydrogen, we can use Graham's law of effusion. According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molecular weight (M).
The formula derived from Graham's law for two gases is:
Rate1 / Rate2 = √(Molecular Weight2 / Molecular Weight1)
Since hydrogen gas (H2) has a molecular weight of approximately 2.016 g/mol, and we know the rate of effusion of the unknown gas is 1/50 that of hydrogen, we can rearrange the formula to solve for the unknown gas's molecular weight (Munknown).
Let's use 'r' to represent the rate of effusion, and 'M' to represent the molecular weight:
rH2 / runknown = √(Munknown / MH2)
Since rH2 / runknown = 50, we can square both sides to get:
(50)2 = (Munknown / 2.016)
Multiplying both sides by 2.016, we get the molecular weight of the unknown gas:
Munknown = (50)2 x 2.016
Munknown ≈ 5040 g/mol
This is an incredibly high molecular weight for a gas; it is possible that the question is asking for a diffusion rate comparison other than 1/50, or this could be a hypothetical scenario. In the context of this calculation, typical molecular weights for gasses are generally under 200 g/mol, so the result should generally be much lower.