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- Frage: (st-16)
Berechne die Molmasse folgender Verbindungen:
- Wasser
- Kohlendioxid
- C6H12O6
- NaCl
Textantwort: - M(H2O): 2·M(H) + 1·M(O) = 2·1+1·16 = 18 g/mol
- M(CO2): 1·M(C) + 2·M(O) = 1·12+2·16= 44 g/mol
- M(C6H12O6): 6·M(C) + 12·M(H) + 6·M(O) = 6+12+12·1+6·16 = 180 g/mol
- M(NaCl): M(Na+) + M(Cl–) = 23.0 +35.5 = 58.5 g/mol
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