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Cm=n/V
n=0,6dm3 x 2
n=1,2 moli substancji
2.
n=0,1 x 200dm3
n=20 moli NaBr
NaBr
masa cz: 23+80=103g/mol
103g - 1 mol
X - 20 moli
X=2060g NaBr