Question

Cine îmi poate rezolva aceasta problema de clasa a 9a???

Answer 1

ms1 = 300g, c1% = 11,96% p1 = 1,08 g/cm3

ms2 = 250g, c2% = 31,4% p2 = 1,23 g/cm3

ms3 = 600g, c3% = 50,5% p3 = 1,40 g/cm3

V.apa = 600 mL p.apa = 1,0 g/cm3

din fiecare solutie vom calcula atat md cat si numarul de moli

stim ca c% = mdx100/ms

=> md1 = c1%xms1/100 = 11,96x300/100 = 35,88 g

   md2 = c2%xms2/100 = 31,4x250/100 = 78,5 g

   md3 = c3%xms/100 = 50,5x600/100 = 303 g

stim ca nr. moli = n = masa/miu, miuH2SO4 = 98 g/mol

=> n1 = 35,88/98 = 0,366 moli

    n2 = 78,5/98 = 0,8 moli

    n3 = 303/98 = 3,092 moli

stim ca p = ms/Vs => Vs = ms/p

=> Vs1 = 300/1,08 = 277,78 cm3 = 0,28 L

    Vs2 = 250/1,23 = 203,3 cm3 = 0,2 L

    Vs3 = 600/1,4 = 428,6 cm3 = 0,43 L

  m.apa = pxV.apa = 1x600 = 600 g

=>

md.tot = md1+md2+md3 = 417,38 g

ms.tot = ms1+ms2+ms3+m.apa = 1750 g

=> c%.finala = md.totx100/ms.tot

= 417,38x100/1750 = 23,85%

Cm.final = n.tot/Vs.tot

n.tot = n1+n2+n3 = 4,258 moli

Vs.tot = Vs1+Vs2+Vs3+V.apa = 1,51 L

=> Cm.final = 4,258/1,51 = 2,82 mol/L