koliko gr MgSO4*7H20 treba rastvoriti u 1dm3 vode da bi se dobio rastvor molalnosti 1mol/kg?
MgSO4*7H20 = MgSO4 + 7H2O n(kh)=n(bs)=1/7n(H2O,kh) kh-kristalohidrat bs-bezvodna so V(H2O)=1 dm3=1000 cm3 m(H2O)=1000 g (zbog gustine 1 g/cm3) b=n(bs)/m(rastvaraca) m(rastvaraca)=(1000+n(H2O,kh)*M(H2O))g=(1000+7n(bs)*18)g=(1+0,126n(bs)) kg 1=n(bs)/(1+0,126n(bs)) n(bs)=1,14 mol n(bs)=n(kh)=1,14 mol m(kh)=n(kh)*M=136,8 g
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MgSO4*7H20 = MgSO4 +
MgSO4*7H20 = MgSO4 + 7H2O
n(kh)=n(bs)=1/7n(H2O,kh)
kh-kristalohidrat
bs-bezvodna so
V(H2O)=1 dm3=1000 cm3 m(H2O)=1000 g (zbog gustine 1 g/cm3)
b=n(bs)/m(rastvaraca)
m(rastvaraca)=(1000+n(H2O,kh)*M(H2O))g=(1000+7n(bs)*18)g=(1+0,126n(bs)) kg
1=n(bs)/(1+0,126n(bs))
n(bs)=1,14 mol
n(bs)=n(kh)=1,14 mol
m(kh)=n(kh)*M=136,8 g
Non scholae, sed vitae discimus!