Question

3- Given 90.0 g C6H1206 a- How many moles of Glucose are present in 90.0 g of glucose?

Answer 1

molar mass = 180.18

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$90.0 \: g \times \frac{1 \: mol}{180.18 \: g} = 0.4995004995 \: mol$

$= 0.500 \: mol$