The half life of 15 P^ 32 is 14.3 days. A solution is prepared with 175 mg of Na
ID: 2263099 • Letter: T
Question
The half life of 15P^32 is 14.3 days. A solution is prepared with 175 mg of Na3PO4 containing 40% by weight of P as 15P^32. Calculate the activity (in Ci) of the solution after 10 days. Take the atomic weight (g mol-1) of Na = 23, P = 32, and O =16. NAv = 6.022 x 1023 atoms mol-1, 1 Ci = 3.7 x 1010 Bq. 1 day = 86400 s
Hint: Calculate the atoms (No) of 15P^32 in the original solution. The 15P^32 activity (Ao) in the original solution is Ao = kNo and A(t = 10 days) = Aoe-kt . Convert to Ci.
Answer: 2390 Ci
(15 is subscript 32 is superscript for "15P32")
Explanation / Answer
M.W of Na3P04 = 165 g
Given 175 mg of Na3P04
165 g of Na3P04 contains 32 g of Phosphorus
175 mg of Na3p04 contains y g of phosphorus
y= 175 x 32 /165
y= 33.94 mg
Given 40% by weight is 15 P 32
Weight of 15P32 = 33.94 x 40/100
Weight of 15P32 = 13.5757 mg
M.W of 15P32 = 32 g
Moles of phosphorus = weight /M.w
=13.5757 x 10-3 /32
Moles of phosphorus = 0.4242 x 10-3
We know that 1 mole of an element contains 6.022 x 1023 atmos
So 0.4242 x 10-3 moles of an element contains x atoms
x = 6.022 x 1023 x 0.4242 x10-3
x=2.55 x 1020 atoms
Initial activity Ao= kN0
A0= k x
Final activity A = Ao e-kt
Given half life = 14.3 days
Half life = 14.3 x 86400 s
Half life = 1235520 S
k = 0.693 /half life
k= 0.693 / 1235520
t= 10 days
t= 10 x 86400 s
t= 864000 s
kt = 864000 x 0.693/1235520
kt = 0.4846
A= Ao e-0.4846
A= 0.6159 Ao
Ao= kN0
Ao= (0.693 /1235520) x 2.55 x 1020
Ao= 0.143 x 1015
A = 0.6159 Ao
So A= 0.6159 x 0.143 x 1015
A= 0.088 x 1015 Bq
Given 1 ci = 3.7 x 1010 Bq
SO A = 0.088 x 1015 / 3.7 x 1010 ci
A= 2378 ci
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