The amount of sugar `(C_(12)H_(22)O_(11))` required to prepare 22 L of its 0.1 M aqueous solution is:

Options

A. 768.4 g
B. 117.1 g
C. 752.4 g
D. 136.8 g

Explanation

To calculate the amount of sugar (C₁₂H₂₂O₁₁) required to prepare 22 L of a 0.1 M aqueous solution, follow these steps:

### Step 1: Calculate the molecular weight of sugar (C₁₂H₂₂O₁₁)
The molecular weight can be calculated using the atomic weights of carbon (C), hydrogen (H), and oxygen (O):
- Carbon (C): 12 g/mol
- Hydrogen (H): 1 g/mol
- Oxygen (O): 16 g/mol

The molecular formula is C₁₂H₂₂O₁₁, so:
- Weight from Carbon = 12 × 12 = 144 g
- Weight from Hydrogen = 22 × 1 = 22 g
- Weight from Oxygen = 11 × 16 = 176 g

Now, add these together:
\[ \text{Molecular weight} = 144 + 22 + 176 = 342 \text{ g/mol} \]

### Step 2: Calculate the number of moles required
Using the formula for molarity (M):
\[ \text{Molarity (M)} = \frac{\text{Number of moles (n)}}{\text{Volume of solution (L)}} \]

Rearranging gives:
\[ n = M \times \text{Volume} \]

Substituting the values:
\[ n = 0.1 \, \text{mol/L} \times 22 \, \text{L} = 2.2 \, \text{moles} \]

### Step 3: Calculate the weight of sugar required
Using the formula:
\[ \text{Weight} = \text{Number of moles} \times \text{Molecular weight} \]

Substituting the values:
\[ \text{Weight} = 2.2 \, \text{moles} \times 342 \, \text{g/mol} = 752.4 \, \text{g} \]

### Conclusion
The amount of sugar required to prepare 22 L of a 0.1 M aqueous solution is **752.4 grams**.

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