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V=1,5dm³=1500cm³
d=?
d=m÷V
d=425g÷1500cm³
d≈0,28g/cm³
Cn=n÷Vr [mnożnik]
mNaCl=23g+35,5g
mNaCl=58,5g w 1 molu
58,5g - 1 mol
425 g - x mol
_______________
x≈7,26 moli
Cn=7,26÷1,5dm³
Cn=4,84mol/dm³
Odp. Stężenia molowe tego roztworu wynosi 4,84mol/dm³.