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At the turn of 60 grams of grape sugar, x kJ is released. Calculate the ΔHof from this.
Solution: M(Traubenzucker) = M(C6H12O6) = 180 grams
With 60 g C6H12O6 x kJ, at a Mol (=180 g) thus 3·x.
ΔHof(C6H12O6) = 3·x kJ/mol.
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