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 meaning of 17% by mass
A 17.0% by mass solution means that there are 17 grams of sodium acetate (solute) in every 100 grams of the solution.

### Step 2: Calculate the number of moles of sodium acetate
To calculate the number of moles of sodium acetate, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass of solute}}{\text{molar mass of solute}} \]

Given:
- Mass of sodium acetate = 17 g
- Molar mass of sodium acetate (CH₃COONa) = 82.0 g/mol

Now, substituting the values:

\[ \text{Number of moles} = \frac{17 \, \text{g}}{82.0 \, \text{g/mol}} \approx 0.207 \, \text{moles} \]

### Step 3: Calculate the volume of the solution
We know the density of the solution is 1.09 g/mL. To find the volume of the solution, we can use the formula:

\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \]

Rearranging this gives us:

\[ \text{Volume} = \frac{\text{mass}}{\text{density}} \]

The mass of the solution is 100 g (as we are considering 100 g of the solution). Now substituting the values:

\[ \text{Volume} = \frac{100 \, \text{g}}{1.09 \, \text{g/mL}} \approx 91.743 \, \text{mL} \]

### Step 4: Convert volume from mL to L
Since molarity is expressed in moles per liter, we need to convert the volume from milliliters to liters:

\[ \text{Volume in liters} = \frac{91.743 \, \text{mL}}{1000} \approx 0.091743 \, \text{L} \]

### Step 5: Calculate molarity
Now we can calculate the molarity (M) using the formula:

\[ \text{Molarity} (M) = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Substituting the values we have:

\[ M = \frac{0.207 \, \text{moles}}{0.091743 \, \text{L}} \approx 2.257 \, \text{M} \]

Rounding this off gives us:

\[ M \approx 2.26 \, \text{M} \]

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

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