Sporządzic roztwór soli n NaH2PO4 o v=260 cm3 i stęz proc P=1,2 %( gestosc= 1,006 g/cm3 ). Ile UWODNIONEJ soli nalezy odważyc aby sporzadzic ten roztwor?
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Cp=ms/mr * 100%
Cp=1,2%
mr= 1,006g/cm3*260cm3=261,56g
1,2%=ms/261,56g *100%
ms=3,14g
NaH2PO4*2H2O
120gNaH2PO4---36gH2O
3,14gNaH2PO4---xgH2O
x=0,942g H2O
3,14g+0,942g=4,082g
Cp = 1,2%
mr = v/d = 260 cm3/1,006 g/cm3 = 258,44g
ms = mr * Cp/100%
ms = 258,44*1,2/100
ms = 2,15g
M NaH2PO4 = 120g/mol
120g - 36g H2O
2,15g - x g
x = 0,645g
2,15 + 0,645 = 2,795