Answer to Question #110347 in General Chemistry for Thomas Velez

Question

What mass of C6H12O6 (glucose) is needed to prepare 450. mL of a 0.650 M solution of glucose in water

Answer 1

m(C6H12O6)=Mr(C6H12O6)*C(C6H12O6)*V(C6H12O6)=180*0.65*0.45=52.65 g

Answer: 52.65 g of C6H12O6 (glucose) is needed to prepare 450. mL of a 0.650 M solution of glucose in water