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

To find the amount of glucose required to prepare 250 mL of \( \frac{M}{20} \) aqueous solution, we need to use the concept of molarity, which is defined as the number of moles of solute per liter of solution. Here is the step-by-step explanation: 

1. The molarity of the glucose solution is given as \( \frac{M}{20} \). This means: \(M = \frac{1}{20}\) mol L^-1.
2. Calculate the number of moles of glucose required for 250 mL (or 0.25 L) of this solution: \(n = \text{Molarity} \times \text{Volume in Liters}\). 
   \(n = \frac{1}{20} \times 0.25 = \frac{1}{80} \, \text{mol}\).
3. Now, convert the moles of glucose to grams using the molar mass of glucose (180 g mol^-1): \(\text{mass} = n \times \text{Molar Mass} = \frac{1}{80} \times 180\).
4. Calculate the mass: \(\text{mass} = \frac{180}{80} = 2.25 \, \text{grams}\).

Therefore, the amount of glucose required to prepare 250 mL of \( \frac{M}{20} \) aqueous solution is 2.25 grams, which corresponds to the correct option.

This value matches the correct answer provided: 2.25 g.