For 1.8 mole of the radioactive isotope xenon-133, half-life = 5.3 days, how many disintegrations have

Question

For 1.8 mole of the radioactive isotope xenon-133, half-life = 5.3 days, how many disintegrations have occurred after 20 minutes?

Answer 1

m = 1.8(0.5)^t/5.3
when m is in moles and t is in days. The time is given in minutes, so 5.3 days = 5.3*24*60, and
m = 1.8(0.5)^20/(5.3*24*60)
m = 1.7967334054851011088942949831 moles
n = N(1.8 - m) where N is Avogadro's Number, 6.02214179*10^23 atoms/mole
n = 6.02214179*10^23(1.8 - 1.796733405485101108894294983115)
n ≈ 1,967,189,533,915,738,975,232 or 1.96719*10^21 disintegrations