Question

Polonium-210, 210Po, decays to lead-206, 206Pb, by alpha emission according to the equation 210 Po 206Pb + He 84 If the half-life, t 112, of 210 Po is 138.4 days, calculate the mass of 206 Pb produced from a 562.0 mg sample of polonium(IV) chloride, PoC14, that is left untouched for 334.1 days. Assume that the only polonium isotope present in the sample is the 210 Po isotope. The isotopic molar masses of 210 Po is 209.98 g/mol and 206Pb is 205.97 g/mol. mass of 206Pb: mg

Answer 1

Solution: Mass of ²⁰⁶Po produced = 61.9106 mg

Explanation:

Given data:

• Half life (t_1/2) of ²¹⁰Po = 138.4 days
• Mass of PoCl₄ sample = 562.0 mg
• Molar mass of ²¹⁰Po = 209.98 g/mol
• Molar mass of ²⁰⁶Pb = 205.97 g/mol

Molar mass of PoCl4 ( Assuming that only ²¹⁰Po isotopes of Polonium are present in PoCl₄)

= (Atomic mass of ²¹⁰Po) + 4 * (Atomic mass of Cl)

= ( 209.98 ) g + ( 4 * 35.453) g

= 351.792 g

Step 1: Calculation of Mass of ²¹⁰Po in 562.0 mg sample of PoCl₄

Percentage mass of ²¹⁰Po in PoCl₄ = ( Molecular mass of ²¹⁰Po ) / ( Molar mass of PoCl₄ ) * 100

= { ( 209.98 g) / (351.792 g) } * 100

= ( 0.59688 ) * 100

= 59.69 %

Therefore, Mass of ²¹⁰Po in 562.0 mg sample of PoCl₄ = 59.69 % of 562.0 mg

= 0.5969 * 562.0 mg

= 335.4503 mg

Step 2: Calculation of Mass of ²¹⁰Po left after t = 334.1 days

This is the inital amount of ²¹⁰Po present. Let this be denoted as A_o

Hence, A_o = 335.4503 mg

Since, radioactive decay follows first-order kinetics, hence

Half-life ( t_1/2) of ²¹⁰Po = ln 2 / k

where k is the decay constant.

t_1/2 = ln 2 / k

or k = ln2 / t_1/2

or k = ln2 / ( 138.4 days ) [ Given,  t_1/2 = 138.4 days ]

or k = 0.693 / 138.4

or k = 0.005 days^-1

Using the integrated rate expression for first order reaction,

A = A_oe^-kt

where A is the amount of ²¹⁰Po left after time t = 334.1 days

A_o is the initial amount of ²¹⁰Po = 335.4503 mg

k is the decay constant, k = 0.005/day

Hence, A = ( 335.4503 mg ) e^{- {  ( 0.005/day ) * 334.1 days ) }}

  A = ( 335.4503 mg ) e^{- ( 1.6705 )}

A =  ( 335.4503 mg ) * ( 0.1881529 )

A = 63.11596 mg

This is the amount of ²¹⁰Po left after t = 334.1 days

No. of moles of ²¹⁰Po in 63.1159 mg = 63.11596 mg / Molar mass of ²¹⁰Po

= ( 63.11596 * 10^-3 g ) / ( 209.98 g )

= 0.3006 * 10^-3 moles of  ²¹⁰Po

Step 3: Calculation of mass of ²⁰⁶Pb produced

Now from the alpha decay equation of ²¹⁰Po:

²¹⁰Po   $\rightarrow$ ²⁰⁶Pb +  ₂⁴He

1 mole of ²¹⁰Po produces 1 mole of ²⁰⁶Pb.

Hence, 0.3005 * 10^-3 moles of  ²¹⁰Po will produce = 0.3006 * 10^-3 moles of  ²⁰⁶Pb .

Mass of one mole of ²⁰⁶Pb = 205.97 g

Therefore, Mass of 0.3005 * 10^-3 moles of  ²⁰⁶Pb = 0.3006 * 10^-3  * 205.97 g

= 61.9106 * 10^-3 g

= 61.9106 mg

Hence, Mass of ²⁰⁶Pb produced = 61.9106 mg (Ans.)