Helium diffuses four times faster than which | 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_1/r_2 = \s

Vedclass · Accepted Answer

$SO_2$

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 $M$: $r_1/r_2 = \sqrt{M_2/M_1}$.Given $r_{He}/r_x = 4$,where $M_{He} = 4 \ g/mol$.Substituting the values: $4 = \sqrt{M_x/4}$.Squaring both sides: $16 = M_x/4$.Therefore,$M_x = 16 	imes 4 = 64 \ g/mol$.Comparing the molar masses of the options:$M(ClO_2) = 35.5 + 2(16) = 67.5 \ g/mol$$M(SO_2) = 32 + 2(16) = 64 \ g/mol$$M(CO_2) = 12 + 2(16) = 44 \ g/mol$$M(NO_2) = 14 + 2(16) = 46 \ g/mol$The gas with a molar mass of $64 \ g/mol$ is $SO_2$.

Helium diffuses four times faster than which of the following?

Similar Questions

At $27 \ ^oC$,the ratio of $rms$ velocities of ozone to oxygen is

Two flasks $X$ and $Y$ have capacities of $1 \ L$ and $2 \ L$ respectively,and each contains $1 \ mol$ of gas. The temperature of the flasks is maintained such that the average speed of molecules in flask $X$ is twice that in flask $Y$. The pressure in flask $X$ is ...

The temperature of the gas is raised from $27\,^oC$ to $927\,^oC$. The root mean square velocity is:

The root mean square velocity of one mole of a monoatomic gas having molar mass $M$ is $V_{rms}$. The relation between the average kinetic energy $(E)$ and the $V_{rms}$ is

Explain the effect of the mass of a molecule on the Maxwell-Boltzmann speed distribution.

Vedclass Test Series

Exam Paper Generator

Online Exam Module