The ratio of number of oxygen atoms (O) in 16.0g ozone (O), 28.0 g carbon monoxide (CO) and 32.0g oxygen (O2) is :(Atomic mass :C =12, O =16 and Avogadro’s constant NA = 6.0×1023 mol-1)

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3 : 1 : 1

1 : 1 : 2

3 : 1 : 2

1 : 1 : 1

answer is B.

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Detailed Solution

Molar mass of O3 = 48Given 16 g O3 . So no. of moles of O3 = 16/48 = ⅓1 mole = 3 ×NA oxygen atomsSo 1/3 mole = NA×3×1/3 no of atoms= NA oxygen atomsMolar mass of CO = 28Given 28 g CO. So no of moles = 28/28 = 1No. of atoms = 1×NA = NAMolar mass of O2 = 32Given 32g O2No. of moles = 32/32 = 1No.of atoms = 2×NA = 2NASo the ratio is 1:1:2

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