How Many Grams of Glucose (C₆H₁₂O₆,180 $\times$ 10² G
B)1 $\times$ 10² G
C)5

How many grams of glucose (C₆H₁₂O₆,180.2 g/mol) would be required to prepare 2.50 L of a 0.278 M glucose solution that could be used in an IV bag?

A) 6.95 $\times$ 10² g
B) 1.11 $\times$ 10² g
C) 5.01 $\times$ 10¹ g
D) 4.50 $\times$ 10² g
E) 1.25 $\times$ 10² g