To find out how many grams of potassium carbonate (K₂CO₃) are needed to make 400.0 mL of a 2.5 M solution, we follow these steps:
Step 1: Convert the volume from milliliters (mL) to liters (L).
Since there are 1000 mL in a liter:
[tex]\[ 400.0 \, \text{mL} = \frac{400.0}{1000} \, \text{L} = 0.400 \, \text{L} \][/tex]
Step 2: Use the molarity formula to find the number of moles of K₂CO₃ required.
The molarity (M) formula is:
[tex]\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \][/tex]
We can rearrange this to solve for moles of solute:
[tex]\[ \text{moles of K₂CO₃} = \text{Molarity} \times \text{liters of solution} \][/tex]
[tex]\[ \text{moles of K₂CO₃} = 2.5 \, \text{M} \times 0.400 \, \text{L} = 1.0 \, \text{moles} \][/tex]
Step 3: Use the molar mass of K₂CO₃ to convert moles to grams.
Given that the molar mass of K₂CO₃ is 138.21 g/mol:
[tex]\[ \text{grams of K₂CO₃} = \text{moles of K₂CO₃} \times \text{molar mass of K₂CO₃} \][/tex]
[tex]\[ \text{grams of K₂CO₃} = 1.0 \, \text{moles} \times 138.21 \, \text{g/mol} \][/tex]
[tex]\[ \text{grams of K₂CO₃} = 138.21 \, \text{g} \][/tex]
Therefore, the required amount of potassium carbonate to make 400.0 mL of a 2.5 M solution is 138.21 grams.
So, the correct answer is not listed among the options provided. The closest numerical value provided in the options is D. [tex]\(140 \, \text{g K₂CO₃}\)[/tex], but it is not the exact answer. Therefore, there appears to be an error in the provided options. The accurate answer is [tex]\(138.21 \, \text{g}\)[/tex] and none of the provided options match exactly.
