If 27.0 mL of AgNO3 is needed to precipitate all the CI^-1 ions in a 0.795-mg sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution?
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To find the molarity of the AgNO3 solution, we need to first determine the number of moles of AgNO3 used.
1. Convert the mass of KCl to moles:
Given mass of KCl = 0.795 mg
First, we convert the mass to grams:
0.795 mg = 0.795 × 10^(-3) g = 7.95 × 10^(-4) g
Next, we use the molar mass of KCl (39.10 g/mol) to convert grams to moles:
Moles of KCl = (7.95 × 10^(-4) g) / (39.10 g/mol) = 2.03 × 10^(-5) mol
2. From the balanced chemical equation, we know that 1 mol of AgNO3 reacts with 1 mol of KCl to form 1 mol of AgCl.
Therefore, the number of moles of AgNO3 used is also 2.03 × 10^(-5) mol.
3. Now, we can calculate the molarity of the AgNO3 solution.
Molarity (M) = Moles of solute / Volume of solution (in liters)
Given volume of AgNO3 solution = 27.0 mL = 27.0 × 10^(-3) L
Molarity of AgNO3 = (2.03 × 10^(-5) mol) / (27.0 × 10^(-3) L) = 0.751 M
Therefore, the molarity of the AgNO3 solution is 0.751 M.
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