Answer: The percent by volume of [tex]SO_2[/tex] in air is 0.00005%
Explanation:
We are given:
0.5 ppm by volume of [tex]SO_2[/tex] in air. This means that [tex]0.5\mu L[/tex] of sulfur dioxide is present in 1 L of solution (air).
To calculate the percent by volume of sulfur dioxide in air, we use the equation:
[tex]\%\text{ volume of sulfur dioxide in air}=\frac{\text{Volume of }SO_2}{\text{Volume of air}}\times 100[/tex]
We are given:
Volume of [tex]SO_2=0.5\muL=0.5\times 10^{-6}L[/tex] (Conversion factor: [tex]1L=10^{6}\mu L[/tex] )
Volume of air = 1 L
Putting values in above equation, we get:
[tex]\%\text{ volume of sulfur dioxide in air}=\frac{0.5\times 10^{-6}\mu L}{1L}\times 100\\\\\%\text{ volume of sulfur dioxide in air}=0.00005\%[/tex]
Hence, the percent by volume of [tex]SO_2[/tex] in air is 0.00005%
