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