Question

An adult of 65 kg drinks water for 5 years contaminated with toluene at $0.15$ mg/L. Reference dose (RfD) of toluene $=0.200$ mg/(kg$ . $d). Daily water intake $=2$ L/d. Compute the hazard quotient (HQ) for the adult (rounded to three decimals).

Hint:

For non-carcinogenic water ingestion, the exposure duration cancels in ADD if $AT=ED\times 365$ and $EF=365$ d/y. Always check units: mg/L $\times$ L/d $\Rightarrow$ mg/d; then divide by kg.

Answer 1

Correct Answer: 0.022

Step 1: Average daily dose (non-carcinogenic).
Use \[ \text{ADD}=\frac{C\,(\text{mg/L})\times IR\,(\text{L/d})\times EF\,(\text{d/y})\times ED\,(\text{y})}{BW\,(\text{kg})\times AT\,(\text{d})}. \] For non-cancer, $AT=ED\times 365$ and typically $EF=365$, so $EF\times ED/AT=1$. Therefore, \[ \text{ADD}=\frac{C\times IR}{BW}. \] Plug values: $C=0.15$ mg/L, $IR=2$ L/d, $BW=65$ kg, hence \[ \text{ADD}=\frac{0.15\times 2}{65}=0.004615\ \text{mg/(kg$ . $d)}. \]
Step 2: Hazard quotient.
\[ HQ=\frac{\text{ADD}}{\text{RfD}}=\frac{0.004615}{0.200}=0.023075\ \Rightarrow\ \boxed{0.023}. \] (An HQ $<1$ indicates exposure below the reference threshold.) Final Answer: \(\fbox{0.023}\)