To determine the volume of a 2.5 M solution that can be prepared using 300 grams of potassium carbonate ([tex]\( K_2CO_3 \)[/tex]), we can follow these steps:
1. Calculate the number of moles of potassium carbonate ([tex]\( K_2CO_3 \)[/tex]):
- First, we need to know the molar mass (molecular weight) of potassium carbonate, which is given as 138 grams per mole (g/mol).
- Given the mass of potassium carbonate is 300 grams.
- Using the formula:
[tex]\[
\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}}
\][/tex]
Substituting in the values:
[tex]\[
\text{Number of moles of } K_2CO_3 = \frac{300 \text{ g}}{138 \text{ g/mol}} \approx 2.173913 \, \text{moles}
\][/tex]
2. Calculate the volume of the 2.5 M solution:
- Molarity (M) is defined as the number of moles of solute per liter of solution ([tex]\(\text{M} = \frac{\text{moles}}{\text{liters}}\)[/tex]).
- We need to find the volume in liters that can be prepared with the given molarity of 2.5 M.
- Rearranging the formula for molarity to solve for volume:
[tex]\[
\text{Volume} = \frac{\text{Number of moles}}{\text{Molarity}}
\][/tex]
Substituting in the values:
[tex]\[
\text{Volume of the solution} = \frac{2.173913 \, \text{moles}}{2.5 \, \text{M}} \approx 0.869565 \, \text{liters}
\][/tex]
Therefore, you can prepare approximately 0.869565 liters (or about 870 milliliters) of a 2.5 M potassium carbonate solution using 300 grams of potassium carbonate.
