The number of sulphur atoms in 3.20g of SO\(_{2(g)}\) is [O = 16.0;S = 32.0; Avogadro constant = 6.02 x 10\(^{23}\)]
Answer Details
To determine the number of sulfur atoms in 3.20g of SO\(_{2(g)}\), we first need to find the number of moles of SO\(_{2(g)}\) present in the given mass. This can be done using the molecular weight of SO\(_{2(g)}\), which is calculated by adding the atomic weights of sulfur and oxygen in the molecule:
Molecular weight of SO\(_{2(g)}\) = (1 × atomic weight of S) + (2 × atomic weight of O)
= (1 × 32.0) + (2 × 16.0)
= 64.0 g/mol
Next, we can calculate the number of moles of SO\(_{2(g)}\) present in 3.20g using the formula:
Number of moles = Given mass ÷ Molecular weight
Number of moles of SO\(_{2(g)}\) = 3.20 g ÷ 64.0 g/mol
= 0.05 mol
Finally, we can use Avogadro's constant to calculate the number of sulfur atoms present in 0.05 mol of SO\(_{2(g)}\):
Number of sulfur atoms = Number of moles of SO\(_{2(g)}\) × Avogadro's constant × Number of sulfur atoms per molecule
From the balanced chemical equation for the combustion of sulfur dioxide, we know that each molecule of SO\(_{2(g)}\) contains one atom of sulfur. Therefore, the number of sulfur atoms per molecule of SO\(_{2(g)}\) is 1.
Number of sulfur atoms = 0.05 mol × 6.02 x 10\(^{23}\) mol\(^{-1}\) × 1
= 3.01 x 10\(^{22}\)
Therefore, the number of sulfur atoms in 3.20g of SO\(_{2(g)}\) is 3.01 x 10\(^{22}\), and the correct option is (A) 3.01 x 10\(^{22}\).