Which of the following gives the molarity of a `17.0%` by mass solution of sodium acetate,`CH_(3)COONa(FM=82.0amu)` in water? Given the density is 1.09 g/mol.

Options

A. `2.26xx10^(-6)M`
B. 0.207M
C. 2.07M
D. 2.26M

Answer

To find the molarity of a 17.0% by mass solution of sodium acetate (CH₃COONa) in water, we can follow these steps:

### Step 1: Understand the given data
- **Percent by mass**: 17.0% means that in 100 grams of the solution, there are 17 grams of sodium acetate (solute) and 83 grams of water (solvent).
- **Formula mass (molar mass) of sodium acetate (CH₃COONa)**: 82.0 g/mol
- **Density of the solution**: 1.09 g/mL

### Step 2: Calculate the volume of the solution
Using the density formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
We can rearrange this to find the volume:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]
Substituting the mass of the solution (100 g) and the density (1.09 g/mL):
\[ \text{Volume} = \frac{100 \, \text{g}}{1.09 \, \text{g/mL}} \approx 91.74 \, \text{mL} \]

### Step 3: Calculate the number of moles of sodium acetate
Using the formula for moles:
\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]
Substituting the mass of sodium acetate (17 g) and its molar mass (82 g/mol):
\[ \text{Moles of CH₃COONa} = \frac{17 \, \text{g}}{82 \, \text{g/mol}} \approx 0.2073 \, \text{mol} \]

### Step 4: Calculate the molarity of the solution
Molarity (M) is defined as:
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]
First, convert the volume from mL to L:
\[ \text{Volume in L} = \frac{91.74 \, \text{mL}}{1000} \approx 0.09174 \, \text{L} \]
Now, substituting the number of moles and the volume in liters:
\[ \text{Molarity} = \frac{0.2073 \, \text{mol}}{0.09174 \, \text{L}} \approx 2.26 \, \text{M} \]

### Final Answer
The molarity of the 17.0% by mass solution of sodium acetate is approximately **2.26 M**.

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