We add excess NaCl solution (58.44 g/mol) to

64 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
1.91 grams.
What is the molarity of the original AgNO3
solution? The formula weight of NaNO3 is
85.00 g/mol.
Answer in units of M.

NaCl + AgNO3 ==> AgCl + NaNO3.

You have 1.91 g AgCl. How many moles is that?
1.91/143.32 = ??

How many moles AgNO3 is that? Since 1 mole AgCl is produced for every mole AgNO3 initially, moles AgCl must equal moles AgNO3. Then what is the molarity.
M = moles/L. You have moles and you have L. Solve for M.

To find the molarity of the original AgNO3 solution, we need to determine the number of moles of AgNO3 present in the 1.91 grams of AgCl formed.

1. Convert the mass of AgCl to moles:
molar mass of AgCl = 143.32 g/mol
moles of AgCl = mass of AgCl / molar mass of AgCl = 1.91 g / 143.32 g/mol = 0.013334 moles

2. Since AgCl (1:1 ratio) is formed by the reaction between AgNO3 and NaCl, the number of moles of AgNO3 is also 0.013334 moles.

3. Calculate the volume of the AgNO3 solution:
volume of AgNO3 solution = 64 mL = 0.064 L

4. Finally, calculate the molarity:
Molarity (M) = Moles of solute / Volume of solution in liters
Molarity = 0.013334 moles / 0.064 L = 0.208 M

Therefore, the molarity of the original AgNO3 solution is 0.208 M.

To find the molarity of the original AgNO3 solution, we can use the concept of stoichiometry and the information given in the question.

1. Calculate the moles of AgCl formed:
The molar mass of AgCl (silver chloride) is given as 143.32 g/mol. The mass of AgCl obtained is 1.91 grams. We can find the moles of AgCl by dividing the mass by the molar mass:
moles of AgCl = mass of AgCl / molar mass of AgCl

moles of AgCl = 1.91 g / 143.32 g/mol

2. Use stoichiometry to find the moles of AgNO3:
From the balanced chemical equation, we know that 1 mole of AgNO3 reacts with 1 mole of AgCl. Therefore, the moles of AgNO3 will be the same as the moles of AgCl.

moles of AgNO3 = moles of AgCl

3. Calculate the volume of the AgNO3 solution:
The volume of the AgNO3 solution is given as 64 mL.

4. Convert the volume to liters:
volume in liters = volume in mL / 1000

5. Calculate the molarity of the AgNO3 solution:
Molarity (M) = moles of solute / volume of solution in liters

Molarity of AgNO3 = moles of AgNO3 / volume of AgNO3 solution in liters

Now, let's plug in the values:

Molarity of AgNO3 = moles of AgNO3 / (64 mL / 1000)

Molarity of AgNO3 = moles of AgCl / (0.064 L) (converting 64 mL to liters)

Since the moles of AgNO3 and AgCl are the same:

Molarity of AgNO3 = moles of AgCl / (0.064 L)

Molarity of AgNO3 = (1.91 g / 143.32 g/mol) / (0.064 L)

Molarity of AgNO3 ≈ 0.022 M

Therefore, the molarity of the original AgNO3 solution is approximately 0.022 M.

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