400 mL of a 0.5 M aqueous ammonia solution is heated. During the process, some ammonia gas escapes and the volume of the solution is reduced to 200 mL . If the molarity of the final solution is 0.6 M , the mass of ammonia that escaped is x

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400 mL of a 0.5 M aqueous ammonia solution is heated. During the process, some ammonia gas escapes and the volume of the solution is reduced to 200 mL . If the molarity of the final solution is 0.6 M , the mass of ammonia that escaped is x 10⁻² g . The value of x is. (Molar mass of NH ₃ is 17 g mol ⁻¹ )

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Correct answer: 136

Initial moles of NH ₃ = M ₁ V ₁ = 0.5 0.400 = 0.20 mol Final moles of NH ₃ = M ₂ V ₂ = 0.6 0.200 = 0.12 mol Moles of NH ₃ evaporated = 0.20 - 0.12 = 0.08 mol Mass of NH ₃ evaporated = 0.08 17 = 1.36 g Given mass = x 10⁻² g Therefore, x = 136 . Answer: 136

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