1.

Two gases A and B having the same volume diffuse though a porous partition in 20 and 10 seconds respcetively. The molecular mass of A is 49u. Molecular mass of B will be:(a) 50.00 u(b) 12.25 u(c) 6.50 u(d) 25.00 u

Answer»

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.


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