Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1

The correct option is 15.38g.

To find the mass of sodium acetate (CH₃COONa) required, we can break down the solution into clear steps using the molarity formula.

Step 1: Identify the given values
Molarity of the solution ($M$) = 0.375 mol L^-1
Volume of the solution ($V$) = 500 mL
Molar mass of sodium acetate = 82.0245 g mol^-1

Step 2: Convert the volume from milliliters (mL) to liters (L)
Since molarity is defined in terms of liters of solution, we convert the volume:
$$\mathrm{Volume}=\frac{500\text{ mL}}{1000\text{ mL/L}}=0.500\text{ L}$$

Step 3: Calculate the number of moles of sodium acetate required
Molarity is defined as the number of moles of solute per liter of solution:
$$\mathrm{Molarity}\left(M\right)=\frac{\text{Moles of solute}}{\text{Volume of solution in liters}}$$
Rearranging the formula to solve for the number of moles:
$$\text{Moles of solute}=\mathrm{Molarity}\times \text{Volume in liters}$$
Substituting the given values:
$$\text{Moles of solute}=0.375\text{ mol L}{-1}^{}\times 0.500\text{ L}=0.1875\text{ mol}$$

Step 4: Calculate the mass of sodium acetate required
Now, we relate the moles to the mass using the molar mass:
$$\text{Mass}=\text{Moles}\times \text{Molar mass}$$
Substituting the values:
$$\text{Mass}=0.1875\text{ mol}\times 82.0245\text{ g mol}{-1}^{}$$
$$\text{Mass}\approx 15.38\text{ g}$$

Therefore, the mass of sodium acetate required to prepare the solution is 15.38g.