Question

Ne(g) effuses at a rate that is ______ times that of br2(g) under the same conditions.

Answer 1

From Graham's law of diffusion. Rate of diffusion of Neon/Rate of diffusion of Bromine =âš(Molar mass of bromine gas/Molar mass of Neon) Molar mass of bromine gas =79.904 and Molar mass of Neon = 20.17. So we have. âš(79.904/20.17) from the right hand side of the equation. Hence it follows that. Rate of diffusion of Neon/Rate of diffusion of Bromine = âš3.96 = 1.98. Hence it follows that Rate of Diffusion of Neon is 1.98 or (2 times, to1d.p) the rate of diffusion of Bromine.

Answer 2

Answer : The Ne(g) effuses at a rate that is 2.828 times that of
`Br_2(g)` under the same conditions.

Solution : Given,

Molar mass of neon = 20 g/mole

Molar mass of bromine = 160 g/mole

Rate of diffusion : It is defined as the rate of diffusion is inversely proportional to the square root of the molar mass of the gas.

Formula used :

`Rate\propto √(M)\\(Rate_(Ne(g)))/(Rate_(Br_2(g)))=\sqrt{(M_(Br_2(g)))/(M_(Ne(g)))}`

where,

`Rate_(Ne(g))` = Rate of diffusion of neon gas

`Rate_(Br_2(g))` = Rate of diffusion of bromine gas

`M_(Ne(g))` = Molar mass of neon gas

`M_(Br_2(g))` = Molar mass of bromine gas

Now put all the given values in this formula, we get

`(Rate_(Ne(g)))/(Rate_(Br_2(g)))=\sqrt{(160g/mole)/(20g/mole)}=2.828`

`Rate_(Ne(g))=2.828* Rate_(Br_2(g))`

Therefore, the Ne(g) effuses at a rate that is 2.828 times that of
`Br_2(g)` under the same conditions.