How much mass of sodium acetate is required to make 220 mL of 0.575 molar aqueous solution?

Options

A. 10.45g
B. 11.79g
C. 15g
D. 13g

Explanation

To determine how much mass of sodium acetate is required to make a 220 mL of 0.575 molar aqueous solution, we can follow these steps:

### Step 1: Write the formula for sodium acetate
The formula for sodium acetate is \( \text{C}_2\text{H}_3\text{O}_2\text{Na} \).

### Step 2: Calculate the molar mass of sodium acetate
The molar mass of sodium acetate (C2H3O2Na) can be calculated as follows:
- Carbon (C): 2 atoms × 12.01 g/mol = 24.02 g/mol
- Hydrogen (H): 3 atoms × 1.008 g/mol = 3.024 g/mol
- Oxygen (O): 2 atoms × 16.00 g/mol = 32.00 g/mol
- Sodium (Na): 1 atom × 22.99 g/mol = 22.99 g/mol

Adding these together:
\[ \text{Molar mass of sodium acetate} = 24.02 + 3.024 + 32.00 + 22.99 = 82.034 \text{ g/mol} \approx 82 \text{ g/mol} \]

### Step 3: Convert the volume from mL to L
Given the volume of the solution is 220 mL, we convert this to liters:
\[ 220 \text{ mL} = 220 \times 10^{-3} \text{ L} = 0.220 \text{ L} \]

### Step 4: Use the molarity formula to find the number of moles
The molarity (M) is defined as:
\[ \text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]
Rearranging this gives us:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume} \]
Substituting the values we have:
\[ \text{Number of moles} = 0.575 \text{ mol/L} \times 0.220 \text{ L} = 0.1265 \text{ moles} \]

### Step 5: Calculate the mass of sodium acetate
Using the number of moles and the molar mass, we can find the mass:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
Substituting the values:
\[ \text{Mass} = 0.1265 \text{ moles} \times 82 \text{ g/mol} = 10.39 \text{ g} \]

### Step 6: Round the mass to the appropriate significant figures
Rounding to three significant figures, we find:
\[ \text{Mass} \approx 10.4 \text{ g} \]

### Final Answer
The mass of sodium acetate required to make 220 mL of a 0.575 molar aqueous solution is approximately **10.4 grams**.

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