To determine how many grams of potassium carbonate (K₂CO₃) are needed to make 400.0 mL of a 2.5 M solution, we can use the following steps:
1. Convert the volume from mL to liters:
- Volume [tex]\( V = 400.0 \, \text{mL} \)[/tex]
- Since there are 1000 mL in 1 liter, we convert the volume to liters:
[tex]\[
V = \frac{400.0 \, \text{mL}}{1000} = 0.4 \, \text{L}
\][/tex]
2. Calculate the number of moles of potassium carbonate needed:
- Molarity [tex]\( M = 2.5 \, \text{M} \)[/tex] (moles per liter)
- Number of moles [tex]\( n \)[/tex] is given by the formula:
[tex]\[
n = M \times V = 2.5 \, \text{M} \times 0.4 \, \text{L} = 1.0 \, \text{mol}
\][/tex]
3. Calculate the mass of potassium carbonate required:
- Molar mass [tex]\( MM \)[/tex] of K₂CO₃ is 138.21 g/mol
- The mass [tex]\( m \)[/tex] can be found using the formula:
[tex]\[
m = n \times MM = 1.0 \, \text{mol} \times 138.21 \, \text{g/mol} = 138.21 \, \text{g}
\][/tex]
Thus, to prepare 400.0 mL of a 2.5 M solution of potassium carbonate (K₂CO₃), you will need 138.21 grams of K₂CO₃.
Given the options provided, the closest to our calculated value 138.21 g is:
[tex]\[
140 \, \text{g}
\][/tex]
So, the correct answer is:
140 g K₂CO₃.
