Two gases A and B having the same volume diffuse through a porous in 20 and 10 seconds respectively .The molecular mass of A is 49u.Molecular mass of B will be
Choose any convenient volume size, say 1 L.
rate A = 1L/10 sec
rate B = 1L/20 sec
(rate1/rate2) = sqrt(M2/M1)
Substitute and solve for M2.
25.00u
To find the molecular mass of gas B, we can make use of Graham's law of diffusion. According to Graham's law:
Rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.
Let's denote the rate of diffusion of gas A as R_A and the rate of diffusion of gas B as R_B.
We are given the time taken for each gas to diffuse through the porous, which is 20 seconds for gas A (t_A) and 10 seconds for gas B (t_B). Since the volume is the same for both gases, we can assume that the distance traveled by each gas is the same.
Using Graham's law, we can set up the following proportion:
R_A / R_B = √(M_B / M_A)
where M_A and M_B are the molecular masses of gases A and B, respectively.
Rearranging the equation, we have:
M_B = M_A * (R_A / R_B)^2
Substituting the given values, M_A = 49u, and t_A = 20 seconds, t_B = 10 seconds, we can solve for M_B.
M_B = 49u * (20 / 10)^2
M_B = 49u * (2)^2
M_B = 49u * 4
M_B = 196u
Therefore, the molecular mass of gas B is 196u.