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\(^{22}\).