Question

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:

Answer 1

114.3

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:

`C=2(L)/(day)*8(mg)/(L) *(1)/(70\ kg)\\C=0.22857(mg)/(kg-day)`

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

`H=(0.22857)/(2*10^(-3))\\H=114.3`

The hazard ratio is most nearly 114.3.