Question

Suppose 1.85 g of sodium iodide is dissolved in 100. mL of a 64,0 m M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the sodium iodide is dissolved in it. Round your answer to 2 significant digits.

Answer 1

199) SOLD: NaI '#' Agno 1991 . Agils) † Nawozl94) mass of NQI = 1.85 gram mass molas of Nal = 150 grams moles of Nal = 1.85 = 1.2X10-2 moles 150 molar mass of Aginda = 120 gram Imoles nool given: 64.0 mm, 100 ml moles of Agnoza 64x60-35 6.4X10-3 moles. 10 Nallaq,) + Agnozlaq,) + Agils) t NaNO3(aq) I s 1.2*10-2 6-4810-3... C::: 10% -6:4X10% -6.4x10-3 Eqm:..NaI moles 1.2x10-2-6-4X1073 - 5.6810-3 moles. Volume = 100ml Molarity z moles x 1000 - 5.6X1075 X 1000 volume.. 100 - 2 568103 M - 56mm final molarity of Nal = 56 mm

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