What is the ratio of the rate of diffusion of | vedclass

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

Question

(A) 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 \propt

Vedclass · Accepted Answer

$1 : 4$

Vedclass · Answer

(A) 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 $O_2$ to $H_2$ is given by: $\frac{r_{O_2}}{r_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}}$.The molar mass of $O_2$ is $32 \ g/mol$ and the molar mass of $H_2$ is $2 \ g/mol$.Substituting these values: $\frac{r_{O_2}}{r_{H_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.Thus,the ratio is $1 : 4$.

What is the ratio of the rate of diffusion of oxygen $(O_2)$ to hydrogen $(H_2)$?

Similar Questions

The densities of hydrogen and oxygen are $0.09 \ g \ L^{-1}$ and $1.44 \ g \ L^{-1}$ respectively. If the rate of diffusion of hydrogen is $1$,then the rate of diffusion of oxygen in the same units will be:

The rate of diffusion of a gas (molar mass = $128$) compared to oxygen is:

The normal boiling point of water at $760 \, mm$ is $373 \, K$. The vapor pressure of water at $298 \, K$ is $23 \, mm$. If the enthalpy of vaporization is $40.656 \, kJ/mol$,then its boiling point at $23 \, mm$ pressure will be ........... $K$.

What will be the pressure at $0\,^oC$ in Kolkata,Chennai,and Mumbai (in $,bar$)?

The rate of diffusion of a gas is proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module