Question

CO (Carbon Mono Oxide) as per Indian standard is 13.0 ppm. Write this standard in $mg/m^{3}$ at 298 K and 1 atm.

• A. \( 8~mg/m^{3} \)
• B. \( 10~mg/m^{3} \)
• C. \( 15~mg/m^{3} \)
• D. \( 21~mg/m^{3} \)

Hint:

At standard ambient temperature and pressure (298.15 K and 1 atm), the molar volume of an ideal gas is approximately 24.45 L/mol. Using the simplified formula \( mg/m^{3} = \frac{ppm \times M}{24.45} \) can save you significant time during the exam!

Answer 1

The Correct Option is C

Concept: To convert concentration from parts per million (ppm) to \( mg/m^{3} \) for an ideal gas, we use the following relationship derived from the Ideal Gas Law: \[ Conc. (mg/m^{3}) = \frac{ppm \times \text{Molecular Weight}}{V_{m}} \] Where:
• \( ppm \) is the concentration in parts per million.
• \( Molecular Weight (M) \) is the molar mass of the gas (g/mol).
• \( V_{m} \) is the molar volume of an ideal gas at the given temperature and pressure (L/mol).

Step 1: Determine the Molecular Weight of CO.
The molecular weight of Carbon Monoxide (CO) is calculated from its constituent atoms: \[ M_{CO} = 12 (\text{Carbon}) + 16 (\text{Oxygen}) = 28~g/mol \]

Step 2: Calculate the Molar Volume (\( V_{m} \)) at 298 K and 1 atm.
Using the ideal gas equation \( PV = nRT \), the molar volume \( V_m = \frac{V}{n} = \frac{RT}{P} \): \[ V_m = \frac{0.0821~L\cdot atm/(mol\cdot K) \times 298~K}{1~atm} \approx 24.45~L/mol \]

Step 3: Convert ppm to \( mg/m^{3} \).
Substitute the values into the conversion formula: \[ Conc. = \frac{13.0 \times 28}{24.45} = \frac{364}{24.45} \approx 14.887~mg/m^{3} \] Rounding to the nearest provided option, we get approximately \( 15~mg/m^{3} \).