Under the same conditions the rates of diffusion of two gases are in the ratio 1:4. The ratio of their vapour densities is

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$2 : 1$

$1 : 2$

$1 : 16$

$16 : 1$

answer is C.

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Detailed Solution

As per Graham's Diffusion law, we know that the rate of diffusion of two gases is directly proportional to $\sqrt{\frac{1}{M}}$

Where $M$ represents the molar mass

Molar $mass = 2 \times$ vapour density $(VD)$

Applying the provided constraints to the equation we have:

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} \Rightarrow \frac{1}{4}$

$= \sqrt{\frac{2 \times {VD}_{2}}{2 \times {VD}_{1}}} \Rightarrow \frac{{VD}_{2}}{{VD}_{1}} = \frac{1}{16}$

${VD}_{1} : {VD}_{2} = 16 : 1$

Therefore, the required answer is $16 : 1$ .

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