How much mass of sodium acetate is required to make 300 mL of 0.575 molar aqueous solution?

Options

A. 14g
B. 15g
C. 19g
D. 13g

Explanation

To find out how much mass of sodium acetate is required to make a 300 mL of 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 (\( \text{C}_2\text{H}_3\text{O}_2\text{Na} \)) 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:
\[ \text{Molar mass} = 24.02 + 3.024 + 32.00 + 22.99 = 82.034 \, \text{g/mol} \]
We can round this to approximately 82 g/mol.

### Step 3: Convert the volume of the solution from mL to L
Given that the volume of the solution is 300 mL, we convert this to liters:
\[ 300 \, \text{mL} = 300 \times 10^{-3} \, \text{L} = 0.3 \, \text{L} \]

### Step 4: Use the molarity formula to find the number of moles
The molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in L}} \]
Rearranging this gives us:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume in L} \]
Substituting the values:
\[ \text{Number of moles} = 0.575 \, \text{mol/L} \times 0.3 \, \text{L} = 0.1725 \, \text{mol} \]

### Step 5: Calculate the mass of sodium acetate
Now, we can find the mass of sodium acetate using the formula:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]
Substituting the values:
\[ \text{Mass} = 0.1725 \, \text{mol} \times 82 \, \text{g/mol} \approx 14.145 \, \text{g} \]
Rounding this gives us approximately 14 g.

### Final Answer
The mass of sodium acetate required is approximately **14 grams**.

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