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)

Updated On: Apr 16, 2025
  • 2.25 g
  • 4.5 g
  • 0.44 g
  • 1.125 g
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The Correct Option is A

Solution and Explanation

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:

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

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