The amount of glucose required to prepare 250 mL of M20 aqueous solution is : (Molar mass of glucose : 180 g mol-1)
- 2.25 g
- 4.5 g
- 0.44 g
- 1.125 g
Answer
(1) 2.25 g
Solution
Moles of glucose = M V
= 120 2501000
= 0.0125 mol
Mass = 0.0125 180
= 2.25 g
Answer: (1)