How much mass of sodium acetate is required to make 280 mL of 0.575 molar aqueous solution?

Options

A. 13.37g
B. 16.48g
C. 14g
D. 12g

Explanation

To find the mass of sodium acetate required to make 280 mL of a 0.575 molar aqueous solution, we can follow these steps:

### Step 1: Write down the formula for sodium acetate.
The chemical 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 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.
Since molarity is expressed in moles per liter, we need to convert the volume from milliliters to liters:
\[ 280 \text{ mL} = 280 \times 10^{-3} \text{ L} = 0.280 \text{ L} \]

### Step 4: Use the molarity formula to find the number of moles.
The formula for molarity (M) is given by:
\[ M = \frac{\text{number of moles}}{\text{volume in L}} \]
Rearranging this gives us:
\[ \text{number of moles} = M \times \text{volume in L} \]
Substituting the values:
\[ \text{number of moles} = 0.575 \text{ mol/L} \times 0.280 \text{ L} = 0.161 \text{ moles} \]

### Step 5: Calculate the mass of sodium acetate.
We can find the mass using the formula:
\[ \text{mass} = \text{number of moles} \times \text{molar mass} \]
Substituting the values:
\[ \text{mass} = 0.161 \text{ moles} \times 82 \text{ g/mol} = 13.22 \text{ g} \]

### Step 6: Round the mass to an appropriate number of significant figures.
The final mass of sodium acetate required is approximately:
\[ \text{mass} \approx 13.22 \text{ g} \]

### Final Answer:
The mass of sodium acetate required to make 280 mL of a 0.575 molar aqueous solution is approximately **13.22 grams**.

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