oblicz stężenie procentowe roztworu otrzymanego przez rozpuszczenie 22g CaCl2*6H2O w 100g H2O
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M CaCl2*6H2O = 20+2*35,5+6*18 = 199g/mol
M CaCl2 = 91g
199g----------22%
91g----------x
______________
x= 10g <= ms
10g ---- 100g tzn, że Cp = 10 %
mrozp= 100 g
cp=?
mr= 122 g
mCaCl2=?
cp=(ms/mr)*100%
MCaCl2*6H2O= 219 g
MCaCl2 = 111 g
hydr - subst
219g - 111 g
22g - x
x= 11,15
cp= (11,15/122)*100%
cp= 9,14%