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 $3.0 \times 10 ^ { 5 }$ yr.

Solution

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Calculate the number of moles of chlorine.

$n \ (NaClO_4) = n \ (Cl)$

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

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

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

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

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

$N = 4.388 \ \text{mol} \times 6.022 \times 10^{23} \ \text{mol}^{-1}$

$N = 2.642 \times 10^{20}$

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

$N = 0.296 \times 2.642 \times 10^{20}$

$N = 7.822 \times 10^{19}$

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