Respuesta :
sulfur dioxide diffuses at one quarter the rate of helium :)
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.
[tex]R\propto \sqrt{\frac{1}{M}}[/tex]
or,
[tex]\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}[/tex]
where,
[tex]R_1[/tex] = rate of effusion of sulfur dioxide gas
[tex]R_2[/tex] = rate of effusion of helium gas
[tex]M_1[/tex] = molar mass of sulfur dioxide gas
[tex]M_2[/tex] = molar mass of helium gas
Now put all the given values in the above formula, we get
[tex]\frac{R_1}{R_2}=\sqrt{\frac{4.0g/mole}{64g/mole}}[/tex]
[tex]\frac{R_1}{R_2}=\frac{1}{4}[/tex]
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.
