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\(^{23}\)]
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
32g of sulphur = 6.02 x 10\(^{23}\) atoms
3.2g of sulphur = x ,
x = \(\frac{3.2 \times 6.02 \times 10^{23}}{32}\) = 0.602 x 10\(^{23}\) = 6.02 x 10\(^{22}\) atom
32g of sulphur = 6.02 x 10\(^{23}\) atoms
3.2g of sulphur = x ,
x = \(\frac{3.2 \times 6.02 \times 10^{23}}{32}\) = 0.602 x 10\(^{23}\) = 6.02 x 10\(^{22}\) atom