Amount of glucose (`C_(6)H_(22)O_(6)`) required to prepare 100 mL of 0.1 M solution is:

Answer

To find the amount of glucose (C₆H₁₂O₆) required to prepare 100 mL of a 0.1 M solution, we can follow these steps:

### Step 1: Understand the formula for molarity
Molarity (M) is defined as the number of moles of solute per liter of solution. The formula is:
\[ \text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]

### Step 2: Convert volume from mL to L
We are given the volume of the solution as 100 mL. To convert this to liters:
\[ \text{Volume in liters} = \frac{100 \text{ mL}}{1000} = 0.1 \text{ L} \]

### Step 3: Calculate the number of moles of glucose
We know the molarity (0.1 M) and the volume (0.1 L). We can rearrange the molarity formula to find the number of moles:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume in liters} \]
Substituting the values:
\[ \text{Number of moles} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{moles} \]

### Step 4: Calculate the molecular weight of glucose
The molecular formula of glucose is C₆H₁₂O₆. To find its molecular weight:
- Carbon (C): 12 g/mol × 6 = 72 g/mol
- Hydrogen (H): 1 g/mol × 12 = 12 g/mol
- Oxygen (O): 16 g/mol × 6 = 96 g/mol

Adding these together:
\[ \text{Molecular weight of glucose} = 72 + 12 + 96 = 180 \, \text{g/mol} \]

### Step 5: Calculate the mass of glucose required
Now, we can calculate the mass of glucose required using the number of moles and the molecular weight:
\[ \text{Mass} = \text{Number of moles} \times \text{Molecular weight} \]
Substituting the values:
\[ \text{Mass} = 0.01 \, \text{moles} \times 180 \, \text{g/mol} = 1.8 \, \text{g} \]

### Conclusion
The amount of glucose required to prepare 100 mL of a 0.1 M solution is **1.8 grams**.

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