contestada

As a result of discharges from local dry cleaner, dinitrotoluene concentration in the groundwater is 8 mg/L. RfD for dinitrotoluene is 2.0 x 10-3 mg/kg-day. The average 70 Kg person drinks 2L/day water. The hazard ratio is most nearly:

Respuesta :

pedrocarratore

Answer:

114.3

Step-by-step explanation:

If a 70kg person ingests 2L of water per day containing 8 mg/L of  dinitrotoluene, the concentration of  dinitrotoluene on that person's body is:

[tex]C=2\frac{L}{day}*8\frac{mg}{L} *\frac{1}{70\ kg}\\C=0.22857\frac{mg}{kg-day}[/tex]

The hazard ratio is defined by dividing the intake dosage (C) by the reference dose (RfD)

[tex]H=\frac{0.22857}{2*10^{-3}}\\H=114.3[/tex]

The hazard ratio is most nearly 114.3.

Otras preguntas