50 mL of each gas A and gas B takes 150 and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to Graham's law of effusion,the rate of effusion $r$ is inversely proportional to the square root of the molecular mass $M$: $\frac{r_A}

Vedclass · Accepted Answer

$20.25$

Vedclass · Answer

(A) According to Graham's law of effusion,the rate of effusion $r$ is inversely proportional to the square root of the molecular mass $M$: $\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$.Since $r = \frac{V}{t}$ and the volumes $V$ are equal,$\frac{r_A}{r_B} = \frac{t_B}{t_A}$.Substituting the given values: $\frac{200}{150} = \sqrt{\frac{36}{M_A}}$.Simplifying the ratio: $\frac{4}{3} = \sqrt{\frac{36}{M_A}}$.Squaring both sides: $\frac{16}{9} = \frac{36}{M_A}$.Solving for $M_A$: $M_A = \frac{36 	imes 9}{16} = \frac{324}{16} = 20.25$.

$50 \ mL$ of each gas $A$ and gas $B$ takes $150$ and $200$ seconds respectively for effusing through a pinhole under similar conditions. If the molecular mass of gas $B$ is $36$,the molecular mass of gas $A$ will be:

Similar Questions

$A$ closed vessel contains an equal number of molecules of $N_2$ and $O_2$ at a pressure of $650 \ mm$. If $N_2$ is completely removed,what will be the pressure in the vessel?

Pressure-volume $(pV)$ work done by an ideal gaseous system at constant volume is (where $E$ is internal energy of the system):

For the diffusion of a gas at pressure $P$,the rate of diffusion is expressed by :

$A$ container contains equal masses of $H_2$,$He$,$CO_2$,and $Ne$ at a certain temperature. Which of the following gases exerts minimum partial pressure?

The ratio of rates of diffusion of gases $X$ and $Y$ of molecular weights $36$ and $64$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module