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23 January, 01:33

Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring 12.0 ft*15.0 ft*8.60ft.

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  1. 23 January, 01:59
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    Given:

    Dimensions of the room = 12 ft * 15 ft * 8.60 ft

    To determine:

    The amount of HCN that gives the lethal dose in the room with the given dimensions

    Explanation:

    As per the World Health Organization, the lethal dose of HCN is around 300 ppm

    300 ppm = 300 mg of HCN / kg of inhaled air

    Volume of air = volume of room = 12 * 15 * 8.6 = 1548 ft³

    Now, 1 ft³ = 28316.8 cm³

    Therefore, the calculated volume of air corresponds to:

    1548 * 28316.8 = 4.383 * 10⁷ cm3

    Density of air (at room temperature 25 C) = 0.00118 g/cm3

    Thus mass of air corresponding to the calculated volume is

    Mass = Density * volume = 0.00118 g/cm3 * 4.383 * 10⁷ cm3

    = 5.172*10⁴ g = 51.72 kg

    Lethal amount of HCN corresponding to 51.72 kg of air would be.

    = 51.72 kg air * 300 mg of HCN/1 kg air = 15516 mg

    Ans: Lethal dose of HCN = 15.5 g
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