Dear Student, 13 Suppose oxide of nitrogen be N2OxMolecular weight of oxide= 2×Atomic weight of N+x×Atomic weight of Oand, Molecular weight of oxide=92i e 2×14+16×x=92or, 16x=92-28or, x=4Therefore, formula of oxide=N2O4 92 g of N2O4= 1 mole of N2O4= N(Avogadro's number) molecules of N2O41 molecule of N2O4 has (7×2+8×4)= 46 electronsSo, 1 mole of N2O4= 46 N electronsHence, the correct answer is (2)14 Potash Alum is K2SO4 Al2(SO4)31 mole of potash alum contains N(Avogadro's number) molecules of potash alum1 molecule of potash alum contains 8 species (2K+, 2Al3+, 4SO42-)Hence, 1 mole of potash alum contains = 8N speciesHence, the correct answer is (3)

Please solve 13 & 14

Dear Student,

13 . Suppose oxide of nitrogen be N_{2} O_{x} Molecular weight of oxide = 2 \times Atomic weight of N + x \times Atomic weight of O and, Molecular weight of oxide = 92 i . e . 2 \times 14 + 16 \times x = 92 or, 16 x = 92 - 28 or, x = 4 Therefore, formula of oxide = N_{2} O_{4} 92 g of N_{2} O_{4} = 1 mole of N_{2} O_{4} = N (Avogadro' s number) molecules of N_{2} O_{4} 1 molecule of N_{2} O_{4} has (7 \times 2 + 8 \times 4) = 46 electrons So, 1 mole of N_{2} O_{4} = 46 N electrons Hence, the correct answer is (2) 14 . Potash Alum is K_{2} {SO}_{4} . {Al}_{2} ({SO}_{4})_{3} 1 mole of potash alum contains N (Avogadro' s number) molecules of potash alum 1 molecule of potash alum contains 8 species (2 K^{+}, 2 {Al}^{3 +}, 4 {SO}_{4}^{2 -}) Hence, 1 mole of potash alum contains = 8 N species Hence, the correct answer is (3)

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