Suppose 15.3 g of potassium chloride is dissolved in 250 mL of a 0.30 M aqueous solution of silver nitrate. Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the potassium chloride is dissolved in it. Round your answer to 3 significant digits.
Answer (1)
To solve this problem, we need to find the final molarity of the chloride anion ( C l − ) after potassium chloride is dissolved in the silver nitrate solution. Let's go through the steps one by one.
Find the Moles of Potassium Chloride ( K Cl ) :
Molar mass of potassium chloride ( K Cl ) is approximately 74.55 g/mol.
Calculate the moles of K Cl using the formula: moles of K Cl = 74.55 g/mol 15.3 g ≈ 0.205 mol
Determine the Initial Moles of Chloride Ion ( C l − ) from K Cl :
The dissolution of K Cl in water is represented as: K Cl → K + + C l −
Therefore, the moles of chloride ion from K Cl is the same as the moles of K Cl : 0.205 mol
Consider the Volume of the Solution :
The total volume of the solution is 250 mL, or 0.250 L (since the problem assumes no volume change upon dissolving K Cl ).
Calculate the Molarity of C l − :
Now, find the final molarity of C l − using the formula for molarity: Molarity ( M ) = volume of solution in liters moles of solute
Substituting the values we found: Molarity of C l − = 0.250 L 0.205 mol ≈ 0.820 M
Therefore, the final molarity of the chloride anion in the solution is approximately 0.820 M , rounded to three significant figures.
