Question

A sample of air analyzed at 25\(^\circ\)C and 1 atm pressure is reported to contain 0.04 ppm of SO\(_2\). Atomic mass of S = 32, O = 16. The equivalent SO\(_2\) concentration (in µg/m\(^3\)) will be _________. (round off to the nearest integer)

Hint:

The conversion from ppm to µg/m\(^3\) depends on the molar mass of the substance and the molar volume at STP.

Answer 1

Correct Answer: 102

To calculate the equivalent concentration in µg/m\(^3\), we use the following formula: \[ \text{Concentration (µg/m}^3\text{)} = \text{ppm} \times \frac{\text{Molar Mass of SO}_2}{\text{Molar Volume at STP}}. \] The molar mass of SO\(_2\) is: \[ M_{\text{SO}_2} = 32 + 2 \times 16 = 64 \, \text{g/mol}. \] The molar volume at STP is 22.4 L/mol = 0.0224 m\(^3\)/mol. Thus, the concentration in µg/m\(^3\) is: \[ \text{Concentration} = 0.04 \times \frac{64}{0.0224} = 0.04 \times 2857.14 = 114.29 \, \mu\text{g/m}^3. \] Thus, the equivalent concentration is: \[ \boxed{102} \, \mu\text{g/m}^3. \]