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A gaseous oxide of nitrogen contains 30.4 % of nitrogen by mass. If the vapour density of the gas is 46 , the total number of atoms present in one molecule of the gas is: (Given: Atomic masses of N = 14 amu and O = 16 amu )

Options

  1. A3
  2. B4
  3. C8
  4. D6

Correct answer

D. 6

Step-by-step solution

Mass percentage of nitrogen ( N ) = 30.4 % Mass percentage of oxygen ( O ) = 100 % - 30.4 % = 69.6 % Moles of N = 30.4 14 2.17 Moles of O = 69.6 16 = 4.35 Simplest molar ratio of N to O : N : O = 2.17 2.17 : 4.35 2.17 = 1 : 2 The empirical formula of the gas is NO ₂ . Empirical formula mass = 14 + (2 16) = 46 g mol ⁻¹ Molar mass of the gas = 2 Vapour density = 2 46 = 92 g mol ⁻¹ Multiplying factor n = Molar mass Empirical formula mass = 92 46 = 2 Molecular formula = ( NO ₂)₂ = N ₂ O ₄ Total number of atoms in one m

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