Answer:
269.9 mg of Lead (Pb-206) Â are formed from a sample of 565 mg of PoCl4 after 334.7 days.
Explanation:
From the given data
Half life of Po =t_{1/2}= 138.4 days
The value of K is given as
[tex]K = \frac{0.693}{t_{1/2}} = 0.693 / 138.4 = 0.005 day^{-1}[/tex]
Here,
Mass of PoCl4 =m_{PoCl4}= 565 mg
The moles of PoCl4 are given as
[tex]n=\frac {m_{PoCl_4}}{MM_{PoCl_4}} =\frac {0.565 g}{(210 + 4 \times 35.5)} =\frac {0.565 g}{352 g/mole} =0.0016 moles[/tex]
so initial concentration of Po = N_o=n=0.00161 moles Â
Time = t=334.7 days
K = 0.005 day^-1
Putting in first order rate equation
[tex]ln \frac{N_t}{N_o}= -Kt\\ln \frac{N_t}{0.00161}= - 0.005 \times 334.7\\ln \frac{N_t}{0.00161}= - 1.674\\ \frac{N_t}{0.00161}= e^{- 1.674}\\\frac{N_t}{0.00161}= 0.1875\\N_t = 0.1875 X 0.00161 \\N_t= 0.000302 moles[/tex]
So the total moles after 334.7 days have passed is given as 0.000302.
The moles of Pb formed are given as
[tex]n_{Pb}=N_o-N_t\\n_{Pb}= 0.00161 - 0.000302 \\n_{Pb}= 0.00131 moles[/tex]
And the mass is given as
[tex]m_{Pb} = n_{Pb} \times MM_{Pb}\\m_{Pb} = 0.00131\times 206 \\m_{Pb} = 0.2699 g \\m_{Pb} = 269.9 mg[/tex]
So 269.9 mg of Lead (Pb-206) Â are formed from a sample of 565 mg of PoCl4 after 334.7 days.
