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25-11-2020
Chemistry

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Calculate the molarity of a solution that contains 183.51 grams of lead (II) bromide in 500.0 mL of the solution

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25-11-2020

Answer: 1M

Explanation:

Molarity = mols/L

moles of lead bromide: 183.51/ 367.0 = 0.5 mol

500 ml/ 1000 mL = .5L

.5 mol / .5 L = 1 mol/L = 1M

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