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Calculate the volume in liters of a 4.6 x 10^-5 M mercury iodine solution that contains 500.mg of mercury(II) iodide HgI2. Round your answer to 2 significant digits.

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Answer:

V HgI2 = 23.92 L

Step-by-step explanation:

C HgI2 = 4.6 E-5 M = 4.6 E-5 mol/L

∴ m HgI2 = 500 mg = 0.500 g

∴ Mw HgI2 = 454.4 g/mol

⇒ V HgI2 = 0.500g * ( mol / 454.4 g ) * ( L / 4.6 E-5 mol )

⇒ V HgI2 = 23.92 L

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