A solution is prepared by placing 27.8 g of KCl in a 0.700 L volumetric flask and adding water to dissolve the solid, then filling the flask to the mark. What is the molarity of an AgNO3 solution if 22.1 mL of the KCl solution react exactly with 43.0 mL of the AgNO3 solution?
27.8g KCl/molar mass KCl = approx 0.37
0.37/0.7 = approx 0.53M
AgNO3 + KCl ==> AgCl + KNO3
mols KCl used = M x L = 0.0221 x 0.53 = ?
mols AgNO3 = the same (since the ratio in the balanced equation is 1:1).
Then M AgNO3 = mols AgNO3/L AgNO3.
To solve this problem, we will use the concept of stoichiometry and the equation:
AgNO3 (aq) + KCl (aq) → AgCl (s) + KNO3 (aq)
Step 1: Calculate the number of moles of KCl used.
Moles of KCl = mass / molar mass
Molar mass of KCl = 39.10 g/mol + 35.45 g/mol = 74.55 g/mol
Moles of KCl = 27.8 g / 74.55 g/mol = 0.373 mol
Step 2: Calculate the number of moles of AgNO3 used.
The reaction is 1:1 stoichiometry between AgNO3 and KCl.
Thus, the moles of AgNO3 used will be the same as the moles of KCl used.
Moles of AgNO3 = Moles of KCl = 0.373 mol
Step 3: Calculate the molarity of AgNO3.
Molarity (M) = Moles / Volume
Volume of AgNO3 = 43.0 mL = 0.0430 L
Molarity of AgNO3 = Moles of AgNO3 / Volume of AgNO3 = 0.373 mol / 0.0430 L = 8.67 M
Therefore, the molarity of the AgNO3 solution is 8.67 M.
To find the molarity (M) of the AgNO3 solution, we can use the concept of stoichiometry and the given volume and concentration of the KCl solution.
First, let's calculate the number of moles of KCl in the 27.8 g sample:
1. Determine the molar mass of KCl:
Moles of KCl = Mass of KCl / Molar mass of KCl
= 27.8 g / (39.10 g/mol + 35.45 g/mol)
≈ 0.500 mol
Next, using the balanced chemical equation between KCl and AgNO3, we can determine the molar ratio between them. From the equation:
KCl + AgNO3 → KNO3 + AgCl
We know that 1 mol of KCl reacts with 1 mol of AgNO3.
Given that 22.1 mL of the KCl solution reacts with 43.0 mL of the AgNO3 solution, we need to convert these volumes into their respective molarities using the formula:
Molarity (M) = Moles / Volume (in L)
Now, let's solve for the molarity of AgNO3:
1. Determine the molarity of KCl:
Molarity (KCl) = Moles (KCl) / Volume (KCl solution in L)
= 0.500 mol / 0.700 L
≈ 0.714 M
2. Calculate the moles of KCl in the 22.1 mL:
Moles (KCl) = Molarity (KCl) * Volume (KCl solution in L)
= 0.714 M * (22.1 mL / 1000 mL/L) (convert mL to L)
≈ 0.0158 mol
3. Using the balanced equation, we know that 1 mol of KCl reacts with 1 mol of AgNO3. Therefore, 0.0158 mol of AgNO3 are required.
4. Calculate the volume of AgNO3 solution in L:
Volume (AgNO3 solution in L) = Moles of AgNO3 / Molarity (AgNO3)
= 0.0158 mol / (43.0 mL / 1000 mL/L) (convert mL to L)
≈ 0.368 L
5. Finally, determine the molarity of AgNO3:
Molarity (AgNO3) = Moles (AgNO3) / Volume (AgNO3 solution in L)
= 0.0158 mol / 0.368 L
≈ 0.043 M
Therefore, the molarity of the AgNO3 solution is approximately 0.043 M.