We add excess NaCl solution (58.44 g/mol) to 72 mL of a solution of silver nitrate (AgNO3 169.88 g/mol), to form insoluble solid AgCl. When it has been dried and weighed, the mass of AgCl (143.32 g/mol) is found to be 2.61 grams. What is the molarity of the original AgNO3 solution? The formula weight of NaNO3 is 85.00 g/mol.

Question

Nikolai Harper

Chemistry

23 May, 12:57

We add excess NaCl solution (58.44 g/mol) to 72 mL of a solution of silver nitrate (AgNO3 169.88 g/mol), to form insoluble solid AgCl. When it has been dried and weighed, the mass of AgCl (143.32 g/mol) is found to be 2.61 grams. What is the molarity of the original AgNO3 solution? The formula weight of NaNO3 is 85.00 g/mol.

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Answers (1)

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Britney Daugherty · Answer 1

The molarity of the original AgNO3 solution is 0.2529 M

Explanation:

Step 1: The balanced equation

AgNO3 + NaCL → AgCl + NaNO3

This means that for 1 mole of AgNO3 consumed, 1 mole of NaCL will be consumed, to produce 1 mole of AgCL and 1 mole of NaNO3

Step 2: Calculating moles of AgCl

Moles of AgCl = mass of AgCL / Molar mass of AgCL

moles of AgCl = 2.61 grams / 143.32g/mole

moles of AgCl = 0.01821 moles

Step 3: Calculating moles of AgNO3

Since there is consumed 1 mole of AgNO3 to produce 1 mole of AgCl, this means that if there is produced 0.01821 moles of AgCl, there is consumed also 0.01821 moles of AgNO3

Step 4: Calculating molarity of AgNO3

Molarity of AgNO3 = moles of AgNO3 / volume of AgNO3

Molarity of AgNO3 = 0.01821 moles / 0.072 L = 0.2529 M = 0.2529 mol/L

The molarity of the original AgNO3 solution is 0.2529 M