What mass of potassium hydroxide (KOH, molar mass = 56.1 g moll) is needed to make 250.0 mL of a 0.125M KOH solution? 0.0313g 1.75g 3.13g 3.51g
The correct answer and explanation is :
The correct answer is 1.75 g of KOH.
Explanation:
To determine the mass of potassium hydroxide (KOH) needed to prepare 250.0 mL of a 0.125 M solution, we follow these steps:
- Understand Molarity Formula: Molarity (M) is defined as: M=moles of soluteliters of solutionM = \frac{\text{moles of solute}}{\text{liters of solution}} Rearranging the formula to solve for moles of KOH: moles of KOH=M×liters of solution\text{moles of KOH} = M \times \text{liters of solution}
- Convert Volume to Liters: The given volume is 250.0 mL, which is converted to liters: 250.0 mL×1 L1000 mL=0.2500 L250.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.2500 \text{ L}
- Calculate Moles of KOH: Using the molarity equation: moles of KOH=0.125 M×0.2500 L=0.03125 moles\text{moles of KOH} = 0.125 \text{ M} \times 0.2500 \text{ L} = 0.03125 \text{ moles}
- Convert Moles to Grams: The molar mass of KOH is 56.1 g/mol, so we calculate: mass of KOH=0.03125 moles×56.1 g/mol=1.75 g\text{mass of KOH} = 0.03125 \text{ moles} \times 56.1 \text{ g/mol} = 1.75 \text{ g}
Thus, 1.75 g of KOH is needed.
The image above visually presents the step-by-step calculation for determining the mass of KOH needed to prepare 250.0 mL of a 0.125 M solution. The correct answer is 1.75 g of KOH.