Question

same volume of two gases X and Y diffuses through same pin holes in 15 and 25 seconds respectively under identical conditions .if the Malar mass X is 30 g/ mole then the molar mass of Y will be ​

Answer 1

Answer:

mass of y = 250/3

Let velocity of x = vx

let velocity of y = vy

rate of diffusion of x = vx/tx = vx/15

rate of diffusion of y = vy/ty = vy/20

using gay lussac law of diffusion we can get the answer

Answer 2

Given:

Same volume of two gases X and Y diffuses through same pin holes in 15 and 25 seconds respectively under identical conditions.

To find:

If the Molar mass X is 30 g/ mole then the molar mass of Y will be ?

Calculation:

According to GRAHAM'S LAW OF DIFFUSION, the rate of diffusion will be inversely proportional to the square root of the molar mass of the gas.

$\boxed{ \bf\therefore \: r \propto \: \dfrac{1}{\sqrt{M}} }$

Now, taking corresponding ratios :

$\sf \therefore \: \dfrac{r_{X}}{r_{Y}} = \sqrt{ \dfrac{M_{Y}}{M_{X}} }$

$\sf \implies \: \dfrac{( \frac{V}{t_{X}} )}{( \frac{V}{t_{Y}} )} = \sqrt{ \dfrac{M_{Y}}{M_{X}} }$

$\sf \implies \: \dfrac{t_{Y}}{t_{X}} = \sqrt{ \dfrac{M_{Y}}{M_{X}} }$

$\sf \implies \: \dfrac{25}{15} = \sqrt{ \dfrac{M_{Y}}{30} }$

$\sf \implies \: \dfrac{5}{3} = \sqrt{ \dfrac{M_{Y}}{30} }$

$\sf \implies \: \dfrac{M_{Y}}{30} = \dfrac{25}{9}$

$\sf \implies \: M_{Y} = 83.33 \: g/mol$

So, molar mass of Y is 83.33 g/mol.