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Mechanism of Adsorption

Under similar...

Question

Under similar conditions, 390 cm $^{3}$ of gas A of molar mass $M_{A}$ and 130 cm $^{3}$ of gas B of molar mass $M_{B}$ diffused out through a porous container in the time ratio of 2: 1. The ratio of $M_{A}$ to $M_{B}$ is:

3 : 8

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8 : 3

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4 : 9

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9 : 4

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Solution

The correct option is D 4 : 9
$R \propto \frac{1}{\sqrt{M}}$

$R = \frac{V}{t}$

$∴ \frac{M_{A}}{M_{B}} = {(\frac{R_{B}}{R_{A}})}^{2}$

$= {(\frac{V_{B}}{V_{A}})}^{2} \times {(\frac{t_{A}}{t_{B}})}^{2}$

$= \frac{4}{9}$

Hence, the correct option is $C$

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Under similar conditions, 390 cm3 of gas A of molar mass MA and 130 cm3 of gas B of molar mass MB diffused out through a porous container in the time ratio of 2: 1. The ratio of MA to MB is:

The correct option is D 4 : 9 R∝1√MR=Vt∴MAMB=(RBRA)2=(VBVA)2×(tAtB)2=49Hence, the correct option is C

Under similar conditions, 390 cm $^{3}$ of gas A of molar mass $M_{A}$ and 130 cm $^{3}$ of gas B of molar mass $M_{B}$ diffused out through a porous container in the time ratio of 2: 1. The ratio of $M_{A}$ to $M_{B}$ is:

The correct option is D 4 : 9
$R \propto \frac{1}{\sqrt{M}}$

$R = \frac{V}{t}$

$∴ \frac{M_{A}}{M_{B}} = {(\frac{R_{B}}{R_{A}})}^{2}$

$= {(\frac{V_{B}}{V_{A}})}^{2} \times {(\frac{t_{A}}{t_{B}})}^{2}$

$= \frac{4}{9}$

Hence, the correct option is $C$