Question

What is the volume of a solution with 8.9 M (molarity) and 54 grams of KOH?
Responses

6.06 L
6.06 L

8.58 L
8.58 L

0.01778
0.01778

0.108 L

Answer 1

To find the volume of a solution with a given molarity and mass of solute, we can use the formula:

\[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
\]

First, we need to determine the number of moles of KOH in 54 grams. The molar mass of KOH can be calculated as follows:

- Potassium (K): approximately 39.1 g/mol
- Oxygen (O): approximately 16.0 g/mol
- Hydrogen (H): approximately 1.0 g/mol

Total molar mass of KOH:

\[
\text{Molar mass of KOH} = 39.1 + 16.0 + 1.0 = 56.1 \, \text{g/mol}
\]

Next, we can calculate the number of moles of KOH in 54 grams:

\[
\text{Moles of KOH} = \frac{\text{mass of KOH}}{\text{molar mass of KOH}} = \frac{54 \, \text{g}}{56.1 \, \text{g/mol}} \approx 0.9623 \, \text{moles}
\]

Now that we have the moles of KOH, we can use the molarity to find the volume of the solution:

\[
\text{Volume (L)} = \frac{\text{moles of solute}}{\text{Molarity}} = \frac{0.9623 \, \text{moles}}{8.9 \, \text{M}} \approx 0.1082 \, \text{L}
\]

Finally, rounding this to three significant figures, the answer is:

\[
\approx 0.108 \, \text{L}
\]

Thus, the volume of the solution is approximately **0.108 L**.