Answer:
125.21 g of glucose
Explanation:
From the question given above, the following data were obtained:
Molar mass of C6H12O6 = 180.16 g/mol
Volume of solution = 2.50 L
Molarity of C6H12O6 = 0.278 M
Mass of C6H12O6 =.?
Next, we shall determine the number of mole of C6H12O6 in the solution. This can be obtained as follow:
Volume of solution = 2.50 L
Molarity of C6H12O6 = 0.278 M
Number of mole of C6H12O6 =?
Molarity = mole /Volume
0.278 = mole of C6H12O6 / 2.50
Cross multiply
Mole of C6H12O6 = 0.278 × 2.50
Mole of C6H12O6 = 0.695 mole
Finally, we shall determine the mass of glucose, C6H12O6, required to prepare the solution as follow:
Molar mass of C6H12O6 = 180.16 g/mol
Mole of C6H12O6 = 0.695 mole
Mass of C6H12O6 =.?
Mole = mass /molar mass
0.695 = mass of C6H12O6 / 180.16
Cross multiply
Mass of C6H12O6 = 0.695 × 180.16
Mass of C6H12O6 = 125.21 g
Thus, 125.21 g of glucose, C6H12O6 is needed for the preparation of the solution.
