Question Detail
Question
How many grams of KBr are needed to prepare 250.0 mL of a 0.200 M solution? (K=39, Br=80)
Correct Answer
Option A is the correct answer.
Detailed Explanation
Moles KBr needed = 0.250 L × 0.200 M = 0.0500 mol. Molar mass KBr = 119 g/mol. Mass = 0.0500 × 119 = 5.95 g.
Hint
Use Molarity = moles/volume to find moles, then convert to grams using molar mass.