Question

A 53.8-mg sample of sodium perchlorate contains radioactive chlorine-36 (whose atomic mass is 36.0 amu). If 29.6% of the chlorine atoms in the sample are chlorine-36 and the remainder are naturally occurring nonradioactive chlorine atoms, how many disintegrations per second are produced by this sample? The half-life of chlorine-36 is 3.0×1053.0 \times 10 ^ { 5 } yr.

Solution

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Answered 2 years ago
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Calculate the number of moles of chlorine.

n (NaClO4)=n (Cl)n \ (NaClO_4) = n \ (Cl)

n (Cl)=1×m (NaClO4)Mr (NaClO4)n \ (Cl) = 1 \times \dfrac{m \ (NaClO_4)}{Mr \ (NaClO_4)}

n (Cl)=1×0.0538 g)122.605 g/moln \ (Cl) = 1 \times \dfrac{0.0538 \ \text{g})}{122.605 \ \text{g/mol}}

n (Cl)=4.388 moln \ (Cl) = 4.388 \ \text{mol}

Multiply the number of moles in the sample by Avogadro's number to derive the number of atoms in the sample.

N=n (Cl)×6.022×1023 mol1N = n \ (Cl) \times 6.022 \times 10^{23} \ \text{mol}^{-1}

N=4.388 mol×6.022×1023 mol1N = 4.388 \ \text{mol} \times 6.022 \times 10^{23} \ \text{mol}^{-1}

N=2.642×1020N = 2.642 \times 10^{20}

There are 29.6% of the chlorine-36 atoms in the sample.

N=0.296×2.642×1020N = 0.296 \times 2.642 \times 10^{20}

N=7.822×1019N = 7.822 \times 10^{19}

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