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Chapter 8: Problem 136

The legal limit for human exposure to \(\mathrm{CO}\) in the work place is \(35 \mathrm{ppm}\). Assuming that the density of air is \(1.3 \mathrm{~g} / 1\), how many grams of \(\mathrm{CO}\) are in \(1.0 \mathrm{l}\) of air at the maximum allowable concentration? (a) \(4.55 \times 10^{-5} \mathrm{~g}\) (b) \(3.5 \times 10^{-5} \mathrm{~g}\) (c) \(2.69 \times 10^{-5} \mathrm{~g}\) (d) \(7.2 \times 10^{-5} \mathrm{~g}\)

Short Answer

Expert verified
Answer: (a) \(4.55 \times 10^{-5} \mathrm{~g}\)

Step by step solution

01

Calculate the mass of air in 1.0 l

To find the mass of air in 1.0 l, we can use the density formula. Since we have the density of air, we can find the mass of air by multiplying the density by the volume. Density = mass/volume Mass of air = Density * Volume = 1.3 g/l * 1.0 l = 1.3 g
02

Calculate the mass of CO in 1.0 l of air using the given concentration

The concentration of CO is given as 35 ppm. That means for every 1,000,000 parts of air, there are 35 parts of CO. We can use this to find out the mass of CO in 1.3 g of air. 35 parts of CO per 1,000,000 parts of air Mass of CO in 1.3 g air = 35/1,000,000 * 1.3 g = (35/1,000,000) * 1.3 g ≈ 4.55 × 10^{-5} g
03

Select the correct answer

Comparing our answer from step 2 with the provided options: (a) \(4.55 \times 10^{-5} \mathrm{~g}\) (b) \(3.5 \times 10^{-5} \mathrm{~g}\) (c) \(2.69 \times 10^{-5} \mathrm{~g}\) (d) \(7.2 \times 10^{-5} \mathrm{~g}\) Our calculated answer is 4.55 × 10^{-5}g, which matches option (a). So, the correct answer is: (a) \(4.55 \times 10^{-5} \mathrm{~g}\)

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