Question

Radium-226 (226.025402 amu) decays to Radon-224 with a half life of 1.6x10³ years. What volume of Radon gas (at 25 degrees Celsius and 1.0 atm) is produced by 25.0 grams of Radium in 475.0 days?

Answer 1

The first thing to do is to find the decay constant we can use the next equation

t_1/2 = 0.693 / k, we know from the statement that t_1/2 is the half life time which  is 1600 years so

1600 = 0.693 / k , k is the decay constant

k = 0.693 / 1600 = 0.000433 years^-1

For question b we are asked for some days so lets change the constant from years^-1 to days^-1, to do this we multiply by 1 / 365

K = 0.000433 * (1/ 365) = 0.0000011863 days^-1

We know that N = N_o e^-λt

where N is the ammount of sample available at the end and No is the ammount of sample available at the begining, t is the time provided in the statement so,

N = 25 * e^{-(0.0000011863*475)}

N = 24.986 grams at the end of Radium 226 so the mass of Rn 224 formed is

N = 25 - 24.98 = 0.014 grams of Radium 226 decayed, calculate the number of moles of Radium that are decayed

moles = 0.014 / 226 = 0.0000623 moles of Radium 226 decayed and turned to Rn 224

apply ideal gas law equation

V = n R T / P,

P is pressure

V is volume

R is gas constant : 0.082 atm L / K mol

T is Temperature (Kelvin or Fahrenheit)

V =  0.0000623 * 0.082 * 298.15 / 1 = 0.001525 L