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.

or,
..........(1)
where,
= rate of effusion of nitrogen gas = 
= rate of effusion of sulfur dioxide gas = ?
= molar mass of nitrogen gas = 28 g/mole
= molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:


Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.
1) moles = mass/mR
CaCO3 Mr = 40 + 12 + (16×3)
= 52 + 48
= 100
mass = 15
so the moles would be 15 ÷ 100
which is 0.15 moles of CaCO3
2) moles = mass ÷ Mr
Mr of Al2O3 = 27 + (16×3)
= 27 + 48
= 75
mass = 204
so the moles would be 204/75 which is 2.72 moles of Al2O3