solutie I
Vs1 = 4L = 4000 cm3, c% = 63%
solutie II
Vs2 = 6 L = 6000 cm3, Cm = 2 M
sin sol. I
ro = ms/Vs
=> ms1 = roxVs = 1,53x4000 = 6120 g sol H2SO4
ms2 = roxVs2 = 1,14x6000 = 6840 g sol. H2SO4
=> ms.tot = ms1+ms2 = 12960 g
stim ca c% = mdx100/ms
=> din sol. I avem md1 = ms1xc1%/100
= 6120x63/100 = 3855,6 g H2SO4
din sol. II avem
Cm = md/miuxVs, miu.H2SO4 = 98 g/mol
=> md2 = CmxVsxmiu = 2x98x6 = 1176 g H2DO4
=> md.tot = md1 + md2 = 5031,6 g
=> c%.finala = md.totx100/ms.tot
= 5031,6x100/12960 = 38,8% (rotunjit)
pt. calculul Cm.final
avem Vs1 + Vs2 = 4+6 = 10 L
stim ca nr.moli = masa/miu
niu1 = md1/98 = 3855,6/98 = 39,34 moli
niu2 = md2/98 = 1176/98 = 12 moli
=> niu.tot = miu1+miu2 = 51,34 moli H2SO4
=> Cm.finala = 51,34/10 = 5,134 M
