A-Level The mole and reacting-quantity calculations: worked solution

Question

A semiconductor plant wants to gauge its daily pollution rating from the air it discharges. On a given day the plant emits 18.0 dm³ of phosphine. Using your answer from part (b)(i), find the smallest volume of discharged air needed to keep the phosphine concentration within the allowed limit. (If you did not reach an answer in (b)(i), take the LC50 value as 30.0 ppm.)

Allowed limit (LC50) = 30.0 ppm = 30 / 10⁶ by volume. Required: volume-fraction of phosphine <= 30 x 10⁻⁶. Concentration = V(PH3) / V(total air) = 18.0 / V. Set 18.0 / V = 30 x 10⁻⁶, so V = 18.0 / (30 x 10⁻⁶) = 18.0 / (3.0 x 10⁻⁵) = 6.0 x 10⁵ dm³. This is the minimum total discharge volume (the air needed is essentially the same, since the 18 dm³ of phosphine is negligible against 6 x 10⁵ dm³).