The mass of 35S in the Na₂S₂O₃ sample is approximately 4.70 grams.
How can you find the mass of 35S in the sample based on the decay rate and half-life
The decay rate, 7.3 x 10¹¹ disintegrations per second, represents the number of 35S nuclei decaying per second.
We can assume that this decay rate is proportional to the initial number of 35S nuclei present (N₀). This proportionality can be expressed using the decay constant (λ).
The relationship between decay rate (R), initial number of nuclei (N₀), and decay constant (λ) is given by: R = λ * N₀
The half-life (t_½) is the time it takes for half of the radioactive nuclei to decay.
For 35S, t_½ = 88 days. We can convert this to seconds:
t_½ = 88 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute = 7.584 x 10⁶ seconds
The relationship between decay constant and half-life is:
λ = ln(2) / t½
λ = ln(2) / (7.584 x 10⁶ seconds) ≈ 9.03 x 10⁻⁸ s⁻¹
Substituting the decay rate and decay constant into the decay formula: 7.3 x 10¹¹ disintegrations/second = 9.03 x 10⁻⁸ s⁻¹ * N₀ N₀ ≈ 8.08 x 10¹⁸ nuclei
To find the mass of 35S, we need to know the mass of one 35S nucleus. This can be obtained from the atomic mass of 35S (34.968 u).
The total mass of 35S in the sample is: Mass = N₀ * Mass per nucleus Mass = 8.08 x 10¹⁸ nuclei * 34.968 u/nucleus Mass ≈ 2.82 x 10²⁰ u
1 u is approximately 1.6605 x 10⁻²⁴ grams. Therefore: Mass ≈ 2.82 x 10²⁰ u * 1.6605 x 10⁻²⁴ g/u Mass ≈ 4.70 grams
Therefore, the mass of 35S in the Na₂S₂O₃ sample is approximately 4.70 grams.