20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is M. (Nearest Integer value)

Given: Na = 23

Question

20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is  M. (Nearest Integer value)

Given: Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol

Canvas Classes · Accepted Answer

Strategy:
 Identify the balanced reaction between  and  and use the mass of the precipitate to find the initial concentration.

Step 1: Balanced Equation

Formula mass of .

Step 2: Calculate moles of precipitate

Step 3: Calculate Molarity
Moles of  Moles of .
Volume = 20 mL = 0.020 L.

Nearest integer = 1.

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