Answer:
The percent composition of glucose in the mixture is 85 % (Answer E)
Explanation:
"percent composition of glucose = (mass of glucose / mass of the mixture) x 100
Step 1: calculate the mass of glucose ( C6H12O6)
mass of glucose = moles of glucose x molar mass of glucose
molar mass of C6H12O6 = [( 12 x6) + ( 1 x12) + ( 16 x6)= 180 g/mol
mass is therefore= 1.3 moles x 180g/mol = 234 grams
Step 2: calculate the percent composition
= mass of glucose /mass of mixture x 100
= 234 g/276 g x 100 = 85 %(answer E)"