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Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. Molecular mass of B will be
A
25.00 u
B
50.00 u
C
12.25 u
D
6.50 u
Detailed Solution
Graham's law: $\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$
Rate of diffusion = volume diffused / time taken, so $\frac{r_A}{r_B} = \frac{V_A}{t_A}\times\frac{t_B}{V_B}$
The volumes are equal, so $\frac{r_A}{r_B} = \frac{t_B}{t_A} = \frac{10}{20} = \frac{1}{2}$
$\frac{1}{2} = \sqrt{\frac{M_B}{49}}$
Squaring: $\frac{1}{4} = \frac{M_B}{49}$
$M_B = \frac{49}{4} = 12.25$ u
Rate of diffusion = volume diffused / time taken, so $\frac{r_A}{r_B} = \frac{V_A}{t_A}\times\frac{t_B}{V_B}$
The volumes are equal, so $\frac{r_A}{r_B} = \frac{t_B}{t_A} = \frac{10}{20} = \frac{1}{2}$
$\frac{1}{2} = \sqrt{\frac{M_B}{49}}$
Squaring: $\frac{1}{4} = \frac{M_B}{49}$
$M_B = \frac{49}{4} = 12.25$ u