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A 500 mL potassium chloride solution was prepared by dissolving

potassium chloride in distilled water. 25.0 mL of the solution was
titrated with 0.300 M silver nitrate solution. 28.90 mL of silver
nitrate solution was required to reach the end point in the titration.
What is the number of moles of potassium chloride present in the
500.0 mL solution?

𝐾𝐢𝑙(π‘Žπ‘ž) + 𝐴𝑔𝑁𝑂3(π‘Žπ‘ž) β†’ 𝐴𝑔𝐢𝑙(𝑠) + 𝐾𝑁𝑂3(π‘Žπ‘ž)

moles silver nitrate=.3*.0288

moles KCL= moles silver nitrate=.0864
now number moleKCl in 500ml= .0964*500/25=...

Ah, the joys of titration! Let's calculate the number of moles of potassium chloride present in the solution, shall we?

First, we need to determine the molarity (M) of the silver nitrate solution. From the information given, we know that 28.90 mL of the 0.300 M silver nitrate solution was required to reach the end point.

So, moles of silver nitrate = Molarity Γ— Volume
= 0.300 M Γ— 28.90 mL
= 8.67 mmol

Since the balanced equation tells us that the ratio of moles of potassium chloride (KCl) to moles of silver nitrate (AgNO3) is 1:1, we can conclude that there are also 8.67 mmol of KCl present in the solution.

However, we need to convert millimoles (mmol) to moles to match the given units. Remember, there are 1000 mmol in 1 mole.

So, moles of KCl = 8.67 mmol Γ· 1000
= 0.00867 mol

Ta-da! There are approximately 0.00867 moles of potassium chloride in the 500.0 mL solution.

To find the number of moles of potassium chloride in the 500 mL solution, we can use the equation for the reaction between potassium chloride (KCl) and silver nitrate (AgNO3):

KCl(aq) + AgNO3(aq) β†’ AgCl(s) + KNO3(aq)

First, we need to determine the number of moles of silver nitrate used in the titration. We know that the concentration of the silver nitrate solution is 0.300 M, and the volume used is 28.90 mL:

Moles of AgNO3 = Concentration Γ— Volume
Moles of AgNO3 = 0.300 M Γ— 0.02890 L
Moles of AgNO3 = 0.00867 mol

Next, we need to determine the stoichiometric relationship between silver nitrate and potassium chloride. From the balanced chemical equation, we can see that the molar ratio between AgNO3 and KCl is 1:1. Therefore, the number of moles of potassium chloride is also 0.00867 mol.

Lastly, we need to determine the concentration of potassium chloride in the 500 mL solution. We'll use the equation:

Concentration = Moles / Volume
Concentration of KCl = 0.00867 mol / 0.500 L
Concentration of KCl = 0.0173 M

So, the number of moles of potassium chloride present in the 500.0 mL solution is 0.00867 mol, and the concentration of the solution is 0.0173 M.

To find the number of moles of potassium chloride (KCl) present in the 500 mL solution, we need to use the given volume of silver nitrate (AgNO3) solution used in the titration.

First, let's determine the number of moles of AgNO3 used. We know that the concentration of the AgNO3 solution is 0.300 M, and the volume used is 28.90 mL (or 0.0289 L).

Number of moles of AgNO3 = concentration Γ— volume in liters
= 0.300 M Γ— 0.0289 L

Next, we need to determine the mole ratio between KCl and AgNO3. From the balanced chemical equation, we can see that the ratio is 1:1, meaning 1 mole of KCl reacts with 1 mole of AgNO3.

Therefore, the number of moles of KCl present in the solution is equal to the number of moles of AgNO3 used in the titration.

Number of moles of KCl = Number of moles of AgNO3
= 0.300 M Γ— 0.0289 L

Now that we have the number of moles of KCl, we can calculate it using the given information.

First, convert the volume of the 500 mL solution to liters:

Volume of solution = 500 mL = 0.500 L

Now, using the equation:

Number of moles of KCl = Number of moles of AgNO3 Γ— (moles of KCl / moles of AgNO3)

Since the mole ratio between KCl and AgNO3 is 1:1, the expression simplifies to:

Number of moles of KCl = Number of moles of AgNO3

Plugging in the values, we have:

Number of moles of KCl = 0.300 M Γ— 0.0289 L

Calculate this expression to find the number of moles of KCl present in the 500 mL solution.