Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.
Solution :
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] ..........(1)
where,
[tex]R_1[/tex] = rate of effusion of nitrogen gas = [tex]79mL/s[/tex]
[tex]R_2[/tex] = rate of effusion of sulfur dioxide gas = ?
[tex]M_1[/tex] = molar mass of nitrogen gas = 28 g/mole
[tex]M_2[/tex] = molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:
[tex](\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}[/tex]
[tex]R_2=52mL/s[/tex]
Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.