Answer:
There are 0,106 mol of glucose in 19,1g.
Explanation:
First we calculate the weight of 1 mole of Glucose whose molecular formula is C6 H12 0 6, with the atomic weights of each element obtained from the periodic table. Then, by means of a simple rule of three we calculate the moles that are in 19.1 grams:
Weight 1 mol C6H1206 = (Weight C)x 6 + (Weight H)x12 + (Weight 0)x6=
= 12 g x 6 + 1g x12 + 16 g x 6= 180 g/mol
180 g ----1 mol glucose
19,1 g-----x= (19,1 g x 1 mol glucose)/ 180 g= 0,106 mol
