If 26.0mL of AgNO 3 is needed to precipitate all the Cl −ions in a 0.750-mg sample of KCl (forming AgCl), what is the molarity of the AgNO 3 solution
Well, you've stumbled upon a chemical precipitation problem! Looks like you're trying to find the molarity of the AgNO3 solution.
So, let's start by converting the mass of KCl into moles. Given that the molar mass of KCl is 74.55 g/mol, we can calculate:
0.750 mg = 0.750 × 10^-3 g
Now, let's convert the mass of KCl to moles:
moles of KCl = (0.750 × 10^-3 g) / (74.55 g/mol)
Now that we know how many moles of KCl we have, we can use the balanced chemical equation to find the ratio between AgNO3 and Cl-, which is 1:1. From the problem, we know that all the Cl- ions will react.
So, the moles of AgNO3 required are equal to the moles of KCl.
Now, we need to find the volume of AgNO3 solution required. We know that the solution requires 26.0 mL of AgNO3.
So, the molarity (M) of AgNO3 can be calculated by:
M = moles of AgNO3 / volume of AgNO3 (in liters)
Now, I'll leave you with the calculations. Remember to convert milligrams to grams and milliliters to liters in order to ensure consistent units! And don't worry, I'll be here to help with any other questions you have along the way.
To determine the molarity of the AgNO3 solution, we can use the following formula:
Molarity (M) = moles of solute / volume of solution in liters
First, let's calculate the moles of Cl- ions in the 0.750-mg sample of KCl:
Molar mass of KCl = 39.10 g/mol + 35.45 g/mol = 74.55 g/mol
Moles of KCl = mass of KCl / molar mass of KCl
Moles of KCl = 0.750 mg / 74.55 g/mol
Now, let's calculate the moles of AgNO3 needed to react with the moles of Cl- ions:
1 mole of AgNO3 reacts with 1 mole of Cl- ions
Moles of AgNO3 = Moles of Cl- ions
Since the volume of AgNO3 solution required to precipitate all the Cl- ions is 26.0 mL, let's convert it to liters:
Volume of AgNO3 solution = 26.0 mL / 1000 mL/L
Finally, let's calculate the molarity of the AgNO3 solution:
Molarity = Moles of AgNO3 / Volume of AgNO3 solution
Please provide the molar mass of AgNO3 for further calculations.
To find the molarity (M) of the AgNO3 solution, we need to know the number of moles of AgNO3 used and the volume of the solution.
Step 1: Convert the mass of KCl to moles.
Given that the mass of KCl is 0.750 mg, we need to convert it to grams by dividing it by 1000.
0.750 mg = 0.750/1000 = 0.000750 g
Next, we need to convert grams of KCl to moles. To do this, we divide the mass by the molar mass of KCl, which is 74.55 g/mol.
0.000750 g / 74.55 g/mol = 0.00001007 mol
Step 2: Determine the moles of AgNO3 used.
According to the balanced chemical equation between AgNO3 and KCl, 1 mole of AgNO3 reacts with 1 mole of KCl to form 1 mole of AgCl. Therefore, the moles of AgNO3 used will be equal to the moles of KCl.
Hence, the moles of AgNO3 used = 0.00001007 mol.
Step 3: Calculate the volume (in liters) of the AgNO3 used.
We are given that 26.0 mL of AgNO3 is used. To determine the volume in liters, we need to divide the volume by 1000.
26.0 mL = 26.0/1000 = 0.026 L
Step 4: Calculate the molarity (M).
The molarity (M) is defined as the moles of solute divided by the volume of the solution in liters.
Molarity (M) = moles of AgNO3 / volume of AgNO3 solution in liters
Molarity (M) = 0.00001007 mol / 0.026 L
Now, divide the moles of AgNO3 by the volume of the solution to find the molarity of the AgNO3 solution.
Molarity (M) = 0.00001007 mol / 0.026 L ≈ 0.000387 M
Therefore, the molarity of the AgNO3 solution is approximately 0.000387 M.
AgNO3 + KCl ==> AgCl + KNO3
mols KCl = grams/molar mass = about 0.01 mol but you need to do it more accurately.
Convert mol KCl to mol AgNO3. That's 0.01 mol also (look at the coefficients in the balanced equation).
Then M = mols/L. mol = 0.01; L = 0.026
M = 0.01/0.026 = ?