Question

The amount of glucose required to prepare 250 mL of $\frac{M}{20}$ aqueous solution is:
(Molar mass of glucose: 180 g mol-1)

• A. 2.25 g
• B. 4.5 g
• C. 0.44 g
• D. 1.125 g

Answer 1

Answer: (A)

2.25 g

Answer 2

Molarity (M) is defined as moles of solute per liter of solution.

$Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution\ (L)}$

Given that the solution is 1/20 M, the molarity is:

$Molarity = \frac{1}{20} M = 0.05M$

The volume of the solution is 250 mL, which is equal to 0.25 L. We can calculate the moles of glucose needed:

$Moles = Molarity × Volume = 0.05M × 0.25L = 0.0125 moles$

Now, we can calculate the mass of glucose required:

$Mass = Moles × Molar\ mass = 0.0125 moles × \frac{180\ g}{mol} = 2.25 g$