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VMariaS [17]
3 years ago
14

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

Chemistry
1 answer:
garik1379 [7]3 years ago
7 0

<u>Given:</u>

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

<u>To determine:</u>

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

<u>Explanation:</u>

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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