Question

Polonium-210, 210Po, decays to lead-206, 206Pb, by alpha emission according to the equation.

210Po84 --> 206Pb82 + 4He2

If the half-life, t1/2, of 210Po is 138.4 days, calculate the mass of 206Pb that can be produced from a 574.0-mg sample of polonium(IV) chloride, PoCl4, that is left untouched for 331.4 days. Assume that the only polonium isotope present in the sample is the 210Po isotope. The isotopic molar masses of 210Po is 209.98 g/mol and 206Pb is 205.97 g/mol.

Answer 1

210Po84 --> 206Pb82 + 4He2

given t1/2 = 138.4 days

t1/2 = 0.693/K according to radioactive decay rate equation,

138.4 = 0.693/K

Rate constant K = 0.005007

and from Radioactive decay equation,

K = 2.303/t*log(a0/a)

t = time period

a0 = initial concentration

a = final concetration

0.005007 = 2.303/331.4*log(574/a)

a = remaining 210Po84 = 109.2467 mg

amount of Po reacted = 574 - 109.2467

amount of Po reactd = 464.7533mg

1moles of Po produces 1 mole of Pb

209.98 g of Po produces 205.97 g of Pb

464.7533*10^-3 g of Po produces 205.97/209.98*464.7533*10^-3 g of Pb

                               = 455.87 mg

Amount of Pb produced = 455.87 mg