Asked by anonymous _25
What is the molarity of a solution containing 11.522 grams of KOH in enough water to make 350ML of solution?
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First, we need to calculate the number of moles of KOH in the solution:
moles of KOH = mass / molar mass
molar mass of KOH = 39.1 g/mol (K) + 16.0 g/mol (O) + 1.0 g/mol (H) = 56.1 g/mol
moles of KOH = 11.522 g / 56.1 g/mol = 0.205 moles
Next, we need to calculate the molarity of the solution:
molarity = moles of solute / liters of solution
since we have 350 mL of solution, we need to convert to liters by dividing by 1000:
molarity = 0.205 moles / 0.35 liters = 0.586 M
Therefore, the molarity of the solution containing 11.522 grams of KOH in enough water to make 350 mL of solution is 0.586 M.
moles of KOH = mass / molar mass
molar mass of KOH = 39.1 g/mol (K) + 16.0 g/mol (O) + 1.0 g/mol (H) = 56.1 g/mol
moles of KOH = 11.522 g / 56.1 g/mol = 0.205 moles
Next, we need to calculate the molarity of the solution:
molarity = moles of solute / liters of solution
since we have 350 mL of solution, we need to convert to liters by dividing by 1000:
molarity = 0.205 moles / 0.35 liters = 0.586 M
Therefore, the molarity of the solution containing 11.522 grams of KOH in enough water to make 350 mL of solution is 0.586 M.