Question

How does the rate of effusion of sulfur dioxide, so2, compare to that of helium (he)? (note: the molar masses are so2 = 64 g/mol; he = 4.0 g/mol.)?

Answer 1

sulfur dioxide diffuses at one quarter the rate of helium :)

Answer 2

Answer : The rate of effusion of helium gas is four times of the rate of effusion of sulfur dioxide gas.

Solution : Given,

Molar mass of sulfur dioxide gas = 64 g/mole

Molar mass of helium gas = 4.0 g/mole

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

`R\propto \sqrt{(1)/(M)}`

or,

`(R_1)/(R_2)=\sqrt{(M_2)/(M_1)}`

where,

`R_1` = rate of effusion of sulfur dioxide gas

`R_2` = rate of effusion of helium gas

`M_1` = molar mass of sulfur dioxide gas

`M_2` = molar mass of helium gas

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

`(R_1)/(R_2)=\sqrt{(4.0g/mole)/(64g/mole)}`

`(R_1)/(R_2)=(1)/(4)`

Therefore, from this we conclude that, the rate of effusion of helium gas is four times of the rate of effusion of sulfur dioxide gas.