NTA Abhyas JEE Main2020ChemistryChemical KineticsPractice
84 P o 210 decays with a particle to 82 P b 206 with a half-life of 138.4 days. If 1.0 g of 84 P o 210 is placed in a sealed tube, how much helium will accumulate in 69.2 days. Express the answer in c m 3 at STP.
Options
- A28.21 c m 3
- B31.25 c m 3
- C36.85 c m 3
- D38.47 c m 3
Correct answer
B. 31.25 c m 3
Step-by-step solution
t 1 / 2 = 138.4 days, t = 69.2 day Number of half-lifes n = t t 1 / 2 = 69.2 138.4 = 1 2 Amount of Po left 1 2 after half-life = 1 2 1 / 2 g = 0.707 g Amount of Po used in 1 2 half-life = 1 - 0.707 = 0.293 g Now 84 P o 210 → 82 P b 206 + 2 H e 4 210 g Po on decay will produce = 4g He 0.293 g Po on decay will produce = 4 × 0.293 210 = 5.581 × 10 - 3 g H e Volume of He at STP = 5.581 × 10 - 3 × 22400 4 = 31.25 m L = 31.25 c m 3