Question

If 65.1 g lead acetate was used to create 0.5m Solution, what is the mass of the solvent.
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Answer 1

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Answer 2

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Molar mass of lead (II) acetate = 325.3 g/mol

65.1 g × 1 mole / 325.3 g = 0.200 mols pb (C2H3O2)2

0.50m = 0.200 mols / x kg

x = 0.40 kg

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