You visited us 4 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8 : 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 : 9

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

9 : 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $C$

Suggest Corrections

Similar questions

360 c m^{3}

C H_{4}

gas diffused through a porous membrane is

15

minutes. Under similar conditions,

120 c m^{3}

of another gas diffused in

10

minutes. Find the molar mass of the gas.

350 c m^{3}

C H_{4}

and

175 c m^{3}

of an unknown gas

^{'} A^{'}

diffused in the same time under similar conditions. The molecular mass of gas

^{'} A^{'}

is:

Equal volumes of two gases A and B diffuse through a porous pot in 20s and 10s, respectively .If the molar mass of A be $80 g m o l^{- 1}$ ,Find the molar mass of B.

Q. The pressure of

1 g

of an ideal gas

A

300 K

is found to be 2 bar. When

2 g

of another ideal gas

B

is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Which of the following is correct?

(m_{A}

= mass of gas A;

m_{B}

mass of gas B

)

Q. A hydrocarbon gas diffuses through a porous plug diffuses at the

1 / 6^{t h}

rate of diffusion of

H_{2}

gas under similar conditions of

P

and

T

. The gas is:

Join BYJU'S Learning Program

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 $C$