Question

Polonium-210, a naturally occurring radioisotope, is an α emitter, with t1/2 = 138 d. Assume that a sample of 210^Po with a mass of 0.710 mg was placed in a 230.0 mL flask, which was evacuated, sealed, and allowed to sit undisturbed. What would the pressure be inside the flask (in mmHg) at 20 ∘C after 369 days if all the α particles emitted had become helium atoms? Please answer in mmHg, thank you.

Answer 1

₂₁₀Po⁸⁴ → ₂₀₆Pb⁸²  + ₄He²

N(t) = No (1/2)^(t/t1/2)

where N(t) is the mass of isotope at time t

No is the mass of isotope present initially = 0.710 mg

t1/2 is the half life of isotpe = 138 days

t = 369 days

N(t) = 0.710 mg (1/2)^(369/138) = 0.11126 mg

so 0.710 - 0.11126 = 0.59874 mg = 0.59874 * 10^-3 g of Po-210 has reacted

Molar mass of Po-210 = 210 g/mol

No. of moles of Po-210 reacted = Mass / molar mass = 0.59874 * 10^-3 g / 210 g/mol = 2.85 * 10^-6 moles

No. of moles of He produced = no. of moles of Po reacted = 2.85 * 10^-6 moles

Given

T = 20 C = 293 K

V = 230 ml = 0.23 L

R = 0.08206 L.atm/mol.K

PV = nRT

sub known values

P * 0.23 L = 2.85 * 10^-6 moles * 0.08206 L.atm/mol.K * 293 K

P = 2.98 * 10^-4 atm

1 atm = 760 mmHg

2.98 * 10^-4 atm = 2.98 * 10^-4 * 760 mmHg = 0.2265 mmHg Answer