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.)?

Question

asked May 4, 2019 194k views

2 Answers

Gustavo Cardoso · Answer 1 · 2019-05-11T14:12:35+0000

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.