Question

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?

Answer 1

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, $5.4\times 10^{-6}g$

Answer 2

The study of the chemical and bonds is called chemistry.

The correct answer is  $5.4*10^{-6$

What is a volatile compound?

• Volatile organic compounds are organic chemicals that have a high vapor pressure at room temperature.
• High vapor pressure correlates with a low boiling point, which relates to the number of the sample's molecules in the surrounding air, a trait known as volatility

All the data is given in the question. therefore

• Limited level of formaldehyde is $\frac{0.75}{10^6} *7.2 = 5.4*10^{-6$

Hence, the correct answer to the question is $5.4*10^{-6$.

For more information about the solution, refer to the link:-

https://brainly.com/question/6534096