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 $49 a m u$ . Molecular mass of $^{'} B^{'}$ will be:

25 a m u

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50 a m u

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12.25 a m u

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6.50 a m u

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Solution

The correct option is C $12.25 a m u$
Using Graham's law of diffusion
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{m_{2}}{m_{1}}}$
$\frac{V / 20}{V / 10} = \sqrt{\frac{m_{2}}{49}}$
$\frac{1}{2} = \sqrt{\frac{m_{2}}{49}}$
$m_{2} = 12.25 a m u$

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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 amu. Molecular mass of ′B′ will be:

The correct option is C 12.25 amuUsing Graham's law of diffusionr1r2=√m2m1V/20V/10=√m24912=√m249m2=12.25 amu

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 $49 a m u$ . Molecular mass of $^{'} B^{'}$ will be:

The correct option is C $12.25 a m u$
Using Graham's law of diffusion
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{m_{2}}{m_{1}}}$
$\frac{V / 20}{V / 10} = \sqrt{\frac{m_{2}}{49}}$
$\frac{1}{2} = \sqrt{\frac{m_{2}}{49}}$
$m_{2} = 12.25 a m u$