The amount of suga `(C_(12)H_(22)O_(11))` required to prepare 2L of its 0.1 M aqueous solution is

Options

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

Explanation

To find the amount of sugar (C₁₂H₂₂O₁₁) required to prepare a 2L of 0.1 M aqueous solution, we can follow these steps:

### Step 1: Calculate the number of moles of sugar required
We know that molarity (M) is defined as:
\[ \text{Molarity (M)} = \frac{\text{Number of moles (n)}}{\text{Volume (L)}} \]

Given:
- Molarity (M) = 0.1 M
- Volume (V) = 2 L

We can rearrange the formula to find the number of moles:
\[ n = M \times V \]
\[ n = 0.1 \, \text{mol/L} \times 2 \, \text{L} \]
\[ n = 0.2 \, \text{mol} \]

### Step 2: Calculate the molar mass of sugar (C₁₂H₂₂O₁₁)
To find the molar mass, we need to sum the atomic masses 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:
\[ \text{Molar mass of C}_{12}\text{H}_{22}\text{O}_{11} = 144 + 22 + 176 = 342 \, \text{g/mol} \]

### Step 3: Calculate the mass of sugar required
Now that we have the number of moles and the molar mass, we can calculate the mass of sugar required:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass} = 0.2 \, \text{mol} \times 342 \, \text{g/mol} \]
\[ \text{Mass} = 68.4 \, \text{g} \]

### Final Answer
The amount of sugar required to prepare 2L of a 0.1 M aqueous solution is **68.4 grams**.

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