Rate of diffusion ∝ 1/√M
where, M = molecular weight
For gas A and B -
\(\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}\) ...(i)
for same volume of diffusion -
rate of diffusion ∝ 1/Time taken ...(ii)
from equation (i) and (ii) -
\(\frac{t_B}{t_A}=\sqrt{\frac{M_B}{M_A}}\) ...(iii)
Here, given,
MA = 49u
tA = 20 sec
tB = 10 sec
MB = ?
Using equation (iii) and putting the value of MA, tA and tB we got -
10/20 \(=\sqrt{\frac{M_B}{49u}}\)
⇒ 1/4 \(=\frac{M_B}{49u}\)
⇒ MB = 49u/4
⇒ MB = 12.25 u
Hence, molecular mass of gas B will be 12.25 u.