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Answer :
We start by finding the number of moles of potassium hydroxide ([tex]$\text{KOH}$[/tex]) using the formula
[tex]$$\text{moles of KOH} = \frac{\text{mass of KOH}}{\text{molar mass of KOH}}.$$[/tex]
Given that the mass of [tex]$\text{KOH}$[/tex] is [tex]$18.3 \text{ g}$[/tex] and its molar mass is approximately [tex]$56.11 \text{ g/mol}$[/tex], we have
[tex]$$\text{moles of KOH} = \frac{18.3}{56.11} \approx 0.3261 \text{ moles}.$$[/tex]
Next, we use the definition of molarity:
[tex]$$\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}.$$[/tex]
The problem gives a molarity of [tex]$0.550 \text{ M}$[/tex], so we can rearrange the formula to solve for the volume of the solution:
[tex]$$\text{volume (L)} = \frac{\text{moles of KOH}}{\text{molarity}} = \frac{0.3261}{0.550} \approx 0.5930 \text{ L}.$$[/tex]
Finally, to express the answer in milliliters, we convert liters to milliliters knowing that [tex]$1 \text{ L} = 1000 \text{ mL}$[/tex]:
[tex]$$\text{volume (mL)} = 0.5930 \times 1000 \approx 593 \text{ mL}.$$[/tex]
Thus, the volume of the [tex]$0.550 \text{ M}$[/tex] [tex]$\text{KOH}$[/tex] solution that can be made with [tex]$18.3 \text{ g}$[/tex] of [tex]$\text{KOH}$[/tex] is approximately
[tex]$$\boxed{593 \text{ mL}}.$$[/tex]
[tex]$$\text{moles of KOH} = \frac{\text{mass of KOH}}{\text{molar mass of KOH}}.$$[/tex]
Given that the mass of [tex]$\text{KOH}$[/tex] is [tex]$18.3 \text{ g}$[/tex] and its molar mass is approximately [tex]$56.11 \text{ g/mol}$[/tex], we have
[tex]$$\text{moles of KOH} = \frac{18.3}{56.11} \approx 0.3261 \text{ moles}.$$[/tex]
Next, we use the definition of molarity:
[tex]$$\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}.$$[/tex]
The problem gives a molarity of [tex]$0.550 \text{ M}$[/tex], so we can rearrange the formula to solve for the volume of the solution:
[tex]$$\text{volume (L)} = \frac{\text{moles of KOH}}{\text{molarity}} = \frac{0.3261}{0.550} \approx 0.5930 \text{ L}.$$[/tex]
Finally, to express the answer in milliliters, we convert liters to milliliters knowing that [tex]$1 \text{ L} = 1000 \text{ mL}$[/tex]:
[tex]$$\text{volume (mL)} = 0.5930 \times 1000 \approx 593 \text{ mL}.$$[/tex]
Thus, the volume of the [tex]$0.550 \text{ M}$[/tex] [tex]$\text{KOH}$[/tex] solution that can be made with [tex]$18.3 \text{ g}$[/tex] of [tex]$\text{KOH}$[/tex] is approximately
[tex]$$\boxed{593 \text{ mL}}.$$[/tex]
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