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Mekhanik [1.2K]
3 years ago
13

Formaldehyde is a carcinogenic volatile organic compound with a permissible exposure level of 0.75 ppm. At this level, how many

grams of formaldehyde are permissible in a 6.0-L breath of air having a density of 1.2 kg/m3?
Chemistry
1 answer:
Flauer [41]3 years ago
8 0

Answer : The amount of formaldehyde permissible are, 5.4\times 10^{-6}g

Explanation : Given,

Density of air = 1.2kg/m^3=1.2g/L     (1kg/m^3=1g/L)

First we have to calculate the mass of air.

\text{Mass of air}=\text{Density of air}\times \text{Volume of air}

\text{Mass of air}=1.2g/L\times 6.0L

\text{Mass of air}=7.2g

Now we have to calculate the amount of formaldehyde.

Permissible exposure level of formaldehyde = 0.75 ppm = \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}

Amount of formaldehyde in 7.2 g of formaldehyde = 7.2g\times \frac{0.75g\text{ of formaldehyde}}{10^6g\text{ of air}}

Amount of formaldehyde in 7.2 g of formaldehyde = 5.4\times 10^{-6}g

Thus, the amount of formaldehyde permissible are, 5.4\times 10^{-6}g

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