Question:medium

The concentration of chemical A in ambient air is 50 \(\mu g/m^3\) and the toxicological index limit for a threshold health effect for this chemical is 5 mg.

If the average breathing rate is 4 L/min, the minimum time of exposure that can lead to any health effect is ______ days (rounded off to one decimal place).

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Convert the ambient concentration and breathing rate to consistent units (mg/L and L/day), multiply to get the daily inhaled dose, then divide the 5 mg threshold by this daily dose rate to get the exposure time in days.
Updated On: Aug 14, 2026
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Correct Answer: 17.4

Solution and Explanation

An alternative and slightly quicker route is to work entirely in micrograms and minutes first, and convert to days only at the very end.

Convert the breathing rate to m$^3$/min so it matches the units of concentration: $4 \text{ L/min} = 0.004 \text{ m}^3/\text{min}$.

The rate at which chemical A is inhaled is then:
\[ \text{dose rate} = 50\ \mu g/m^3 \times 0.004\ m^3/\text{min} = 0.2\ \mu g/\text{min} \]

The threshold dose of 5 mg equals $5 \times 1000 = 5000\ \mu g$. The time (in minutes) needed to accumulate this much chemical is:
\[ t = \frac{5000\ \mu g}{0.2\ \mu g/\text{min}} = 25000\ \text{min} \]

Converting minutes to days using $1$ day $= 1440$ min:
\[ t = \frac{25000}{1440} = 17.36\ \text{days} \approx 17.4\ \text{days} \]
This agrees exactly with the mg/L-and-per-day calculation, confirming the result.
\[\boxed{t \approx 17.4 \text{ days}}\]
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