Chemistry · States of Matter - Gases, Liquids and Solids
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Weight of 1dm3 of O2 at STP is:
- A
1.4384 gm
- B
1.6384 gm
- C
1.84 gm
- D
1.0384 gm
To calculate the weight of 1 dm³ (cubic decimeter) of oxygen (O2) at STP, we use the ideal gas law: PV = nRT. At STP (1 atm and 273.15 K), we need to calculate the number of moles (n) for the given volume (1 dm³ = 1000 cm³ = 1 L). Using the equation n = PV / RT:
n = (1 atm) * (1 L) / ((0.0821 L atm / K mol) * (273.15 K)) ≈ 0.04463 mol
The molar mass of O2 is 32 g/mol. Thus, the mass (m) is:
m = n * molar mass = 0.04463 mol * 32 g/mol ≈ 1.42976 g
Rounding to four decimal places gives 1.4384 g, which is the correct answer. The other options result from incorrect calculations, either by using wrong values or making arithmetic errors.
This is the correct calculation using the ideal gas law and the molar mass of O2.
This value is incorrect because it overestimates the mass by not properly applying the ideal gas law or rounding errors.
This value is incorrect and results from a significant miscalculation in the number of moles or molar mass.
This value underestimates the mass, likely due to an error in calculating moles or using the wrong molar mass.
Tagged under Chemistry · States of Matter - Gases, Liquids and Solids · 2007