Answer:
k= 80%
Step-by-step explanation:
Jar A contains 4*0.45 L acid, and 4 L of a solution of acid.
Jar B contains 5*0.48 L acid., and 5 L of a solution of acid.
Jar C contains 1*k/100 = k/100 acid, and 1 L of a solution.
50% = 0.5
For jar A.
(2/3)*k/100 L acid is added to jar A.
Now jar A contains 4*0.45 L + (2/3)*k/100 L acid, and it has (4+2/3)L of a solution.
L solute/L solution = 0.5
[4*0.45 L + (2/3)*k/100 L]/(4+2/3)L = 0.5
[1.8 + (2k/300)]/[(12+2)/3] = 0.5
[1.8 + (2k/300)]/[14/3] = 0.5
[1.8 + (2k/300)]= 0.5*(14/3)
(2k/300) = 0.5*(14/3) - 1.8
2k = (0.5*(14/3) - 1.8)*300
k = (0.5*(14/3) - 1.8)*300/2 =80
k= 80%
We also can find k using jar B.
(1/3)k/100 L acid is added to jar B.
Now jar B contains 5*0.48 L+ (1/3)k/100 L acid, and it has (5+1/3) L of a solution.
L solute/L solution = 0.5
[5*0.48 L+ (1/3)k/100 L ]/(5+1/3)L= 0.5
[5*0.48 + (1/3)k/100 ]/(5+1/3)= 0.5
This equation also gives k=80%
Check.
We can check at least for jar A.
Jar A has 4L solution and 4*0.45=1.8 L acid.
2/3 L of the solution from jar C was added, and now we have 4 2/3 L of solution.
(2/3)* 80%= (2/3)*0.8 acid was added from jar C.
Now we have [1.8 +(2/3)*0.8] L acid in jar A.
L solute/L solution = [1.8 +(2/3)*0.8] L /(4 2/3) L = 0.5 or 50% as it is given that jar A has 50% at the end.
is the population parameter that represents the population mean.
