Respuesta :

W0lf93
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 : The Ne(g) effuses at a rate that is 2.828 times that of [tex]Br_2(g)[/tex] 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 :

[tex]Rate\propto \sqrt{M}\\\frac{Rate_{Ne(g)}}{Rate_{Br_2(g)}}=\sqrt{\frac{M_{Br_2(g)}}{M_{Ne(g)}}}[/tex]

where,

[tex]Rate_{Ne(g)}[/tex] = Rate of diffusion of neon gas

[tex]Rate_{Br_2(g)}[/tex] = Rate of diffusion of bromine gas

[tex]M_{Ne(g)}[/tex] = Molar mass of neon gas

[tex]M_{Br_2(g)}[/tex] = Molar mass of bromine gas

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

[tex]\frac{Rate_{Ne(g)}}{Rate_{Br_2(g)}}=\sqrt{\frac{160g/mole}{20g/mole}}=2.828[/tex]

[tex]Rate_{Ne(g)}=2.828\times Rate_{Br_2(g)}[/tex]

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

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