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a
24 g of C (12)
b
56 g of Fe (56)
c
27 g of Al (27)
d
108 g of Ag (108)
answer is A.
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Detailed Solution
(a) 24 g of C (12) has maximum number of atoms.Number of atoms in 24 g of C = 2412 × 6.023 × 1023=2 × 6.023 × 1023 atoms(b) Number of atoms in 56 g of Fe = 5656 × 6.023 × 1023=1 × 6.023 × 1023 atoms(c) Number of atoms in 27 g of Al = 2727 × 6.023 × 1023 atoms=1 × 6.023 × 1023 atoms(d) Number of atoms in 108 g of Ag = 108108 × 6.023 × 1023=1 × 6.023 × 1023 atoms