Under standard conditions of temperature and | vedclass

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

Question

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \f

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.Therefore,the ratio of the rate of diffusion of helium $(He)$ to methane $(CH_4)$ is given by:$\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_{He}}}$Given molar masses: $M_{He} = 4 \ g/mol$ and $M_{CH_4} = 16 \ g/mol$.Substituting the values:$\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2$.

Under standard conditions of temperature and pressure,what will be the ratio of the rate of diffusion of helium to methane?

Similar Questions

The total pressure of a mixture of two gases is equal to .......

If a saturated vapor is slowly compressed (at constant temperature) to half of its initial volume,what happens to the vapor pressure?

$A$ sample of air contains $N_2$ and $O_2$ saturated with water vapor,with a total pressure of $640 \ torr$. If the water vapor pressure is $40 \ torr$ and the molar ratio of $N_2$ to $O_2$ is $3 : 1$,what is the partial pressure of $N_2$ in the sample in $torr$?

$A$ person living in $Shimla$ observed that cooking food without using a pressure cooker takes more time. The reason is that at high altitude:

$50 \ mL$ 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module