What Mass Of Potassium Hydroxide (KOH, Molar Mass – 56.1 G/Mol) Is Needed To Make 100.0 ML Of A 0.125 M KOH Solution? O 7.01 G 2238 O 0.7018 O 0,07018
The correct answer and explanation is:
Correct Answer: 0.701 g
To determine the mass of potassium hydroxide (KOH) required to prepare a 100.0 mL (0.100 L) solution with a concentration of 0.125 M, we use the formula: Mass=Molarity×Volume×Molar Mass\text{Mass} = \text{Molarity} \times \text{Volume} \times \text{Molar Mass}
Given:
- Molarity (M) = 0.125 M
- Volume (V) = 100.0 mL = 0.100 L
- Molar Mass of KOH = 56.1 g/mol
Substituting the values: Mass=(0.125 mol/L)×(0.100 L)×(56.1 g/mol)\text{Mass} = (0.125 \text{ mol/L}) \times (0.100 \text{ L}) \times (56.1 \text{ g/mol}) Mass=0.701 g\text{Mass} = 0.701 \text{ g}
Thus, the correct answer is 0.701 g (rounded from 0.70125 g).
Explanation:
- Understanding Molarity:
Molarity (M) is the number of moles of solute per liter of solution. A 0.125 M solution means there are 0.125 moles of KOH per 1 liter of solution. - Converting Volume:
The problem gives volume in milliliters (100.0 mL), but molarity uses liters, so we convert: 100.0 mL=0.100 L100.0 \text{ mL} = 0.100 \text{ L} - Finding the Number of Moles of KOH:
Using the definition of molarity: Moles of KOH=Molarity×Volume\text{Moles of KOH} = \text{Molarity} \times \text{Volume} =(0.125 mol/L)×(0.100 L)= (0.125 \text{ mol/L}) \times (0.100 \text{ L}) =0.0125 moles= 0.0125 \text{ moles} - Calculating Mass of KOH: Mass=Moles×Molar Mass\text{Mass} = \text{Moles} \times \text{Molar Mass} =(0.0125 moles)×(56.1 g/mol)= (0.0125 \text{ moles}) \times (56.1 \text{ g/mol}) =0.701 g= 0.701 \text{ g}
Conclusion:
To prepare 100.0 mL of 0.125 M KOH solution, you need to weigh 0.701 g of KOH and dissolve it in enough water to make the total volume 100.0 mL.
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