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(d) If 45.3 mL of a 0.108 M HCl solution is needed to

Chapter , Problem 4.81d

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QUESTION:

(d) If 45.3 mL of a 0.108 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

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QUESTION:

(d) If 45.3 mL of a 0.108 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

ANSWER:

Problem 4.81d(d) If 45.3 mL of a 0.108 M HCl solution is needed to neutralize a solution of KOH, how manygrams of KOH must be present in the solution Step-by-step solution Step 1 of 3 (d) To calculate the moles of HCl solution Moles of HCl = Molarity (M) × V olume (M) 1000 ml = 0.108 M × 45.3 ml 1000 ml -3 = 4.89 x 10 mol

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