Doctors use the radioactive isotope chromium-51 to label red blood cells in the human body. Chromium-51 gives off relatively ___
2 answers:
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Answer:
0.269 g
Explanation:
⟶
+
Data:
Half-life of (T(1/2)) = 138.4 days
Mass of PoCl4 = 561 mg (0,561 g) and molecular weight of PoCl4 = 350. 79 g/mol
Time = 338.8 days
Isotopic masses
= 209.98 g/mol
= 205.97 g/mol
Concepts
Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.
The radioactive decay law is
N=Noe^(-λt)
Where:
No = number of atoms in t=0
N = number of atoms in t=t (now) in this case t=338.8 days
λ= radioactive decay constant
The radioactive constant is related to the half-life by the next equation
λ=
so
λ= =0,005008 days^(-1)
No (Atoms of in t=0)
To get No we need to calculate the number of atoms of in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of
in the initial sample.
This is:
= 0,001599 mol of PoCl4
This is the number of mol of in the initial sample.
To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23
0,001599 mol of * 6.023*10^23 atoms/ mol of
= 9.632 *10^20 atoms of
Atoms after 338.8 days
We use the radioactive decay law to get this value
N=Noe^(-λt)
N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20
This is the number of atoms of in the sample after 338.8 days has passed
The number of atoms transformed is equal to the number of atoms of
produced.
The number of atoms of transformed is No - N
9.632 *10^20 – 1.765 *10^20 = 7.866*10^20
So, 7.866*10^20 is the number of atoms of produced
We can get the mass with the Avogadro’s number
(7.866*10^20 atoms of ) / ( 6.023*10^23 atoms of
/ mol of
= 0.001306 moles of
This number of moles have a mass of:
(0,001306 moles of )* (205.97 g of
/mol of
) = 0.269 g
The end product will depend upon
a) the amount of the reagent taken
b) the final treatment of the reaction
If we have just taken methylmagnesium iodide and p-hydroxyacetophenone, then we will get methane and hydroxyl group substituted with MgI in place of hydrogen
Figure 1
However if we have taken excess of methylmagnesium iodide which is Grignard's reagent followed by hydrolysis we will get different product
Figure 2
