Under the same condition of temp.and pressure hydrogen gas diffuses 8 times as fast as gas y.calculate the relative molecular mass of gas y.(R.M
M_2
Thanks, DrBob: the first line should read
8/1=sqrt(MM/2)
Under the same condition of temperature and pressure , hydrogen diffuses 8times as fast as gas Y, calculate the relative molecular mass of Y.(R.M.M of H2= 2)
I can't really get the answer of the question
Ra/Rb=√Mb/Ma
Ra = 8Y
Rb=Y
Ma=2
8Y/Y=√Mb/2
8=√Mb/2
(8)^2=Mb/2
Mb=2×64
Mb=128
To calculate the relative molecular mass of gas Y (RMM₂), we can use Graham's law of effusion.
According to Graham's law, the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be represented as:
Rate₁/Rate₂ = √(Molar mass₂/Molar mass₁)
In this case, we are given that the rate of diffusion of hydrogen gas (Rate₁) is 8 times faster than the rate of diffusion of gas Y (Rate₂). Therefore, the equation becomes:
8/1 = √(Molar mass Y/Molar mass hydrogen)
To find the relative molecular mass of gas Y, we need to square both sides of the equation and isolate the molar mass of Y:
(8/1)² = (Molar mass Y/Molar mass hydrogen)
64 = (Molar mass Y/Molar mass hydrogen)
Now, let's substitute the molar mass of hydrogen with its actual value. The molar mass of hydrogen (H₂) is approximately 2 g/mol.
64 = (Molar mass Y/2)
To find the relative molecular mass of gas Y, we can isolate Molar mass Y:
Molar mass Y = 64 * 2
Molar mass Y = 128 g/mol
Therefore, the relative molecular mass of gas Y (RMM₂) is 128 g/mol.
What is Graham's law? https://www.google.com/search?q=Graham%27s+law&ie=utf-8&oe=utf-8
8/1=sqrt(2/MM)
MM=32 Oxygen O2?
If you mean by RMM compared to H2, then it is 64