Question

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Answer 1

A) Initial concentration given is 0.1 mg/m³ , Expressing it in $\mu$g /m³ we have

1 mg/m³ = 1000 $\mu$g /m³ , Hence 0.1 mg/m³ = 100 $\mu$g /m³

B) 1 ppm = 1 mg/L = 1000 mg/m³ , Hence 1 mg/m³ = 10^-3 ppm

Therefore, 0.1 mg/m³ = 10^-4 ppm

C) Here we would apply the ideal gas equation. The Molecular weight of Formaldehyde = 30 g/mole

PV = nRT, R = PV/nT = (V/n) x (P/T) = constant,

This implies that (V/n)₁ x (P/T)₁ = (V/n)₂ x (P/T)₂ , For the initial limit of 0.1 mg/m³ we have

n/V = 0.1 x 10^-3 / 30 x 1000 L = 10^-7 / 30 moles/Liter,

$\Rightarrow$ (30 / 10^-7) x (1 atm / 298 K) = (V/n)₂ x (0.34 atm / 251 K)

$\Rightarrow$  (V/n)₂ = (251 x 30 / 298 x 0.34) x 10⁷

$\Rightarrow$  (n/V)₂ = 0.0134 x 10^-7 moles/ Liter = 40.36 mg/m³