Question
Ethanethiol (CH3CH2-SH) is a toxic substance and has such a strong odor that a person can detect 0.016 mol dispersed in 5.0×1010 grams of air. Knowing that the density of air is 1.25 g/L and assuming uniform distribution of ethanethiol in the air, the quantity limit, in mol/L, that a person can detect is:
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Answer to a math question Ethanethiol (CH3CH2-SH) is a toxic substance and has such a strong odor that a person can detect 0.016 mol dispersed in 5.0×1010 grams of air. Knowing that the density of air is 1.25 g/L and assuming uniform distribution of ethanethiol in the air, the quantity limit, in mol/L, that a person can detect is:
Frederik
4.6101 Answers
1. Calculate the volume of air using its mass and density:
\text{Volume of air} = \frac{5.0 \times 10^{10} \, \text{g}}{1.25 \, \text{g/L}} = 4.0 \times 10^{10} \, \text{L}
2. Determine the concentration of ethanethiol:
C = \frac{0.016 \, \text{mol}}{4.0 \times 10^{10} \, \text{L}} = 4.0 \times 10^{-13} \, \text{mol/L}
3. The detectable limit of ethanethiol concentration in air:
\boxed{4.0 \times 10^{-13} \, \text{mol/L}}
2. Determine the concentration of ethanethiol:
3. The detectable limit of ethanethiol concentration in air:
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