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OPRA Pharmacokinetics and Pharmacodynamics QUESTION #3830
Question 1
To achieve a chlorine concentration of $1.5$ ppm in a $6000$ L drenching facility, how much chlorinated lime ($30\%\ \text{w/w}$ available chlorine) is needed?
  • $3$ g
  • $27$ g
  • $30$ g✔️
  • $270$ g
Correct Answer Explanation
$1.5\ \text{ppm} = 1.5\ \text{mg/L}$. For $6000$ L: $1.5 \times 6000 = 9000\ \text{mg} = 9\ \text{g}$ chlorine. Since lime is $30\%\ \text{w/w}$: $\dfrac{9}{0.30} = 30\ \text{g}$.