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suppose of potassium chloride is dissolved in of a aqueous solution of ammonium sulfate.calculate the final molarity of potassium cation in the solution. you can assume the volume of the solution doesn't change when the potassium chloride is dissolved in it.be sure your answer has the correct number of significant digits.

User Bechbd
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1 Answer

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The final molarity of potassium cation in the solution is 6.70 × 10⁻⁵ M.

Step 1: Calculate the moles of potassium chloride

moles of potassium chloride = grams of potassium chloride / molar mass of potassium chloride

moles of potassium chloride = 1.75 g / 74.55 g/mol

moles of potassium chloride = 0.0235 mol

Step 2: Calculate the total volume of the solution

total volume of solution = volume of ammonium sulfate solution + volume of potassium chloride solution

total volume of solution = 350.0 mL + 0.0 mL

total volume of solution = 350.0 mL

Step 3: Calculate the final molarity of potassium cation

final molarity of potassium cation = moles of potassium chloride / total volume of solution

final molarity of potassium cation = 0.0235 mol / 350.0 mL

final molarity of potassium cation = 0.000067 M

Round to 3 significant digits:

final molarity of potassium cation = 6.70 × 10⁻⁵ M.

Question

Suppose 1.75g of potassium chloride is dissolved in 350.mL of a 54.0 mM aqueous solution of ammonium sulfate. Calculate the final molarity of potassium cation 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.

User Praneybehl
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