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

The Correct Option is A

Molarity (M) is the concentration of a solution, expressed as moles of solute per liter of solution.

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

For a solution with a concentration of 1/20 M, the molarity is:

\[Molarity = \frac{1}{20} M = 0.05M\]

A solution volume of 250 mL is equivalent to 0.25 L. The required moles of glucose can be calculated as:

\[Moles = Molarity × Volume = 0.05M × 0.25L = 0.0125 moles\]

The mass of glucose required is then determined by:

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