la 70 gr. C --> ms1, md1, c1
la 15 gr.C --> ms2, md2, c2
MCuSO4*5H2O=160+90=250
g/mol
masa
de sare anhidra respectiv de apa din ch depus este :
250g
ch..........160gCuSO4............90gH2O
203,125
ch.........a...........................b a=130g
CuSO4 ; b=73,125 g H2O
in sol . cu
temperatura 15 gr.C (ms2)
ms2 =ms1-203,125
md2=md1-130
aceasta
solutie contine : 1parte de CuSO4/5parti
de H2O =>
c2=1*100/6=100/6=16,66%
sau se mai
poate scrie relatia :
c2/100=(md1-130)/(ms1-203,125)
1/6=(md1-130)/(ms1-203,125)
ms1-203,125=6 md1-780 (1)
am notat x=
nr. moli CuSO4 din ms1 si cu y=nr. moli H2O
din ms1
ms1=mCuSO4+mapa+203,125=160x+18y+203,125 si
md1=160x+130
=> (1) devine :
160x+18y+203,125-203,125=960x+780-780 =>
160x+18y=960x
18y=800x => x=0,0225y
cantitatea
de O din 203,125 g ch este:
250g cristalohidrat......144g O (din
CuSO4+5H2O)
203,125g......................117g O
notez % de O din ms1 cu p%
atunci
in solutia finala (ms2) md2=160x si
ms2=160x+18y
(160x+18y)g sol.............(64x+16y)g O
100g sol............................p%+7,25
p+7,25=(64x+16y) 100/(160x+18y) dar
x=0,0225y
p+7,25= 17,44y*100/21,6y
p+7,25=80,74 ==> p=80,74-7,25=73,49 % aprox. 73,5%
in sol ms1
(160x+18y+203,125)g sol.........(64x+16y+117)gO
100 g sol.........................................73,5
gO
dar x=0,0225y
(17,44y+117)/(21,6y+203,125)=0,7349
17,44y+117=15,87384y+149,276562
32,276562=1,56616y => y=20,6
x = 0,0225*20,6 = 0,4635
pentru solutia initiala ms1
md1=160x+130=74,16+130 = 204,16g CuSO4
ms1=160x+18y+203,125 =74,16+370,8+203,125 = 648,085g solutie
c1=md1*100/ms1=204,16*100/648,085=31,5%
c2=16,66% (s-a calculat la inceput )
