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. 68.4 g
B. 17.1 g
C. 34.2 g
D. 136.8 g

Explanation

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

### Step 1: Understand 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: Calculate the Number of Moles
Given:
- Molarity (M) = 0.1 M
- Volume of solution = 22 L

We can rearrange the formula to find the number of moles:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume} \]
\[ \text{Number of moles} = 0.1 \, \text{M} \times 22 \, \text{L} = 2.2 \, \text{moles} \]

### Step 3: Calculate the Molecular Weight of Sugar (C₁₂H₂₂O₁₁)
To find the molecular weight, we need to sum the atomic weights of all the atoms in the formula:
- Carbon (C): 12 atoms × 12 g/mol = 144 g/mol
- Hydrogen (H): 22 atoms × 1 g/mol = 22 g/mol
- Oxygen (O): 11 atoms × 16 g/mol = 176 g/mol

Now, add these together:
\[ \text{Molecular weight of C}_{12}\text{H}_{22}\text{O}_{11} = 144 + 22 + 176 = 342 \, \text{g/mol} \]

### Step 4: Calculate the Mass of Sugar Required
Now, we can calculate the mass of sugar needed using the number of moles and the molecular weight:
\[ \text{Mass} = \text{Number of moles} \times \text{Molecular weight} \]
\[ \text{Mass} = 2.2 \, \text{moles} \times 342 \, \text{g/mol} = 752.4 \, \text{grams} \]

### Final Answer
The amount of sugar (C₁₂H₂₂O₁₁) required to prepare 22 L of a 0.1 M aqueous solution is **752.4 grams**.

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