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?

A) 3.8 g formaldehyde
B) 5.4 × 10⁻⁶ g formaldehyde
C) 3.8 × 10⁻² g formaldehyde
D) 5.4 g formaldehyde

Option b)

Explanation:

To do this, we know that we have the permissible exposure level of 0.75 ppm, or we can write it as 0.75 mg/kg.

Now, with the quantity of breath air of 6 L, we need to know how much of formaldehyde can be in there. Let's convert the volume of air to mass.

The density of the air is 1.2 kg/m³ or 1.2x10⁻³ kg/L so:

m = 1.2x10⁻³ * 6 = 7.2x10⁻³ kg

Now with this mass, let's see how much formaldehyde is permited in this mass:

mass of formaldehyde = 7.2x10⁻³ kg * 0.75 mg/kg

mass = 0.0054 mg

But the answer is expressed in grams, so let's convert the miligrams to grams:

mass = 0.0054 mg * 1 g/1000 mg = 5.4x10⁻⁶ g

This is the mass and correct answer

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .