A gas, Y, diffuses 1.05 times faster than SO2 gas at the same conditions. What is the molecular weight of gas Y?
difusion rate is inversely proportional to MW. MW of SO2 is 32+32=64
1.05=sqrt (64/M)
M=64/1.05^2
To find the molecular weight of gas Y, we can use Graham's law of effusion, which states:
"The rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass."
Let's assume the molecular weight of SO2 gas is M, and the molecular weight of gas Y is Y.
According to the problem, gas Y diffuses 1.05 times faster than SO2 gas at the same conditions.
This means:
Rate of diffusion of gas Y = 1.05 * Rate of diffusion of SO2 gas
Using Graham's law of effusion, we can set up the following proportion:
(Y / M)^(1/2) = 1.05
Squaring both sides of the equation, we get:
Y / M = 1.05^2
Y / M = 1.1025
Multiplying both sides of the equation by M, we get:
Y = 1.1025 * M
Therefore, the molecular weight of gas Y is 1.1025 times the molecular weight of SO2 gas.
To determine the molecular weight of gas Y, we need to compare its diffusion rate with that of SO2 and use Graham's Law of Diffusion. Graham's law states that the ratio of the rates of diffusion of two gases is inversely proportional to the square root of their molar masses.
Let's assume the molecular weight of SO2 is M. Therefore, the molecular weight of gas Y would be M' (which we need to find).
According to the given information, the diffusion rate of gas Y is 1.05 times faster than SO2. We can express this relation as:
(Diffusion rate of Y) / (Diffusion rate of SO2) = 1.05
By applying Graham's Law, we can rewrite this as:
((Square root of M') / (Square root of M)) = 1.05
Now, let's solve for M'.
First, square both sides of the equation to remove the square roots:
(M' / M) = 1.1025
Now, rearrange the equation to solve for M':
M' = (1.1025) * M
Therefore, the molecular weight of gas Y, M', is 1.1025 times the molecular weight of SO2, M.