A chemist adds 215.0mL of a 6.0x10^−5/mmolL mercury(II) iodide HgI2 solution to a reaction flask. Calculate the micromoles of mercury(II) iodide the chemist has added to the flask. Round your answer to 2 significant digits.

The micromoles of mercury(II) iodide : 0.013 μ moles

Given

215.0mL of a 6.0x10⁻⁵mmol/L HgI₂

Required

micromoles of HgI₂

Solution

Molarity(M) = moles of solute per liters of solution

Can be formulated :

M = n : V

n = moles

V = volume of solution

V = 215 mL = 0.215 L

so moles of solution :

n = M x V

n = 6.10 mmol/L x 0.215 L

n = 1.312 . 10⁻⁵ mmol

mmol = 10³ micromol

so 1.312 mmol = 1.312.10⁻⁵ x 10³ = 0.01312 micromoles ⇒ 2 sif fig = 0.013 μ moles

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