Multiple choice

Equal masses of $SO_{2}$ and $O_{2}$ are kept in a vessel at $27^{o}C$. The total pressure of the mixture is 2.1 $atm$. The partial pressure of $SO_{2}$ is:

  1. 1.4 $atm$
  2. 7 $atm$
  3. 0.7 $atm$
  4. 14 $atm$
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C Correct answer
Explanation

Equal masses mean equal moles of SO2 and O2. Molar mass of SO2 is 64 and O2 is 32. Let mass be m. Moles of SO2 = m/64, moles of O2 = m/32. Mole fraction of SO2 = (m/64) / (m/64 + m/32) = (1/64) / (3/64) = 1/3. Partial pressure = mole fraction * total pressure = (1/3) * 2.1 = 0.7 atm.

AI explanation

Using the mole fraction method for partial pressures, the partial pressure of a gas equals its mole fraction multiplied by the total pressure. For equal masses, assume 64 g of each gas, which results in 1 mole of SO2 and 2 moles of O2. The mole fraction of SO2 is therefore 1 divided by 3, and multiplying this by the total pressure of 2.1 atm gives a partial pressure of 0.7 atm.

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