Question

A sample of Ne pas is observed to effuse through a pour own barrier in 8.61 minutes. Under the same conditions, the same number of moles of an unknown as requires 13.0 minutes to effuse through the same bam The molar mass of the unknown as mol

Answer 1

46.004 g/mol

Step-by-step explanation:

Scottish physicist Thomas Graham formulated a law known as Graham's law of effusion in 1848. He conducted an experiment and found the relationship between the rate of effusion of a gas and its molar mass as:

`r=\sqrt {\frac {1}{M}}`

where,

r is the rate of effusion of a gas

M is the molar mass of the gas.

Also, r = v/t

And for two gases taking different t₁ and t₂ to effuse, the formula is:

`\frac {t_2}{t_1}=\sqrt {\frac {M_2}{M_1}}`

So,

For Neon :

`t_1` = 8.61 minutes

`M_1` = 20.1797 g/mol

For unkown gas:

`t_2` = 13.0 minutes

`M_2` = ? g/mol

`\frac {13.0}{8.61}=\sqrt {\frac {M_2}{20.1797}}`

Molar mass of unknown gas = 46.004 g/mol