The number of turns that citric acid cycle will turn when 57.3 g of glucose (C6H1206) is oxidized in the cells is 1.92 × 10²³ turns.
According to this question, Acetyl CoA turns the citric acid cycle one turn.
First, we find the number of moles of 57.3 grams of glucose by dividing by its molar mass of 180g/mol.
no of moles = 57.3/180 = 0.32moles
Next, we multiply the number of moles by Avogadro's number i.e. 6.02 × 10²³ molecules.
no of turns = 6.02 × 10²³ × 0.32
no of turns = 1.92 × 10²³ turns
Therefore, the number of turns that citric acid cycle will turn when 57.3 g of glucose (C6H1206) is oxidized in the cells is 1.92 × 10²³ turns.
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