Question

A particular air sample at 1 atm pressure has a density of approximately 1.3 g/liter and contains SO2 at a concentration of 25 g/m3. What is this concentration in terms of a. parts per million (mass SO2 per 106 units mass of air)? b. moles SO2 per 106 mol of air?

Answer 1

Ans. #1. Given, [SO₂] = 25 g SO₂ / m³ air = 25 g SO₂ / 1000 L air = 0.025 g SO₂ / L air

Given- density of air = 1.3 g / L

So, mass of 1 L air = volume x density = 1 L x (1.3 g/ L) = 1.3 g

So, [SO₂] = 0.025 g SO₂ / L air = 0.025 g SO₂ / 1.3 g air = 0.019231 g SO₂ / g air

Now, mass of SO₂ in 10⁶ g air = (0.019231 g SO₂ / g air) x 10⁶ g air = 19231 g

Hence, [SO₂] = 19231 g SO₂ / 10⁶ g air = 19231 ppm

#2. Step 1: Calculate moles of air in 1 L air sample (assuming RTP)

#Step 2:

From #1, [SO2] = 0.025 g SO₂ / L air

# From ideal gas equation, 1 L air = 0.0409 mol air

Now, combining above two statements-

            [SO₂] = 0.025 g SO₂ / 0.0409 mol air

                        = (0.025 g / 64.0648 g mol^-1) SO₂ / 0.0409 mol air

                        = 3.9023 x 10^-4 mol SO₂ / 0.0409 mol air

                        = 9.5411 x 10^-3 mol SO₂ / mol air

                        = 9.5411 x 10³ mol SO₂ / 10⁶ mol air

Hence, moles of SO₂ per 10⁶ mol air = 9.5411 x 10³ mol