50 ml of a gas $A$ diffuse through a membrane in the same time as for the diffusion of 40 ml of a gas $B$ under identical pressure temperature conditions. If the Molecular weight of $A = 64$ , that of $B$ would be:

100

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250

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200

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Solution

The correct option is D 80
Rate of diffusion of gas $A = 50 m L = R_{1}$
Rate of diffusion of gas $B = 40 m L = R_{2}$
Mass of gas $A = 64 g = M_{1}$
Mass of gas $B = M_{2} = ?$

From Grahm's law of diffusion,
$R a t e \propto \frac{1}{M o l a r m a s s}$

Formula : $\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$

$∴ \frac{50}{40} = \sqrt{\frac{M_{2}}{64}} \Rightarrow \sqrt{M_{2}} = \frac{5}{4} \times 64 =$

$∴ M_{2} = 80$

Hence answer is (D)

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