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What is the molarity of the resulting solution when 23.640 g of Mn(ClO4)2 · 6 H2O are added to 200.0 mL of water? WebAssign will check your answer for the correct number of significant figures. .326 Incorrect: Your answer is incorrect.

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Answer:

0.3267 M

Explanation:

To solve this problem, first we calculate how many moles of Mn(ClO₄)₂ are contained in 23.640 g of Mn(ClO₄)₂·6H₂O.

Keep in mind that the crystals of Mn(ClO₄)₂ are hydrated, and we need to consider those six water molecules when calculating the molar mass of the crystals.

Molar mass of Mn(ClO₄)₂·6H₂O = 54.94 + (35.45+16*4)*2 + 6*18 = 361.84 g/mol

Now we proceed to calculate:

  • 23.640 g Mn(ClO₄)₂·6H₂O ÷ 361.84 g/mol = 0.0653 mol Mn(ClO₄)₂·6H₂O = mol Mn(ClO₄)₂

Now we divide the moles by the volume, to calculate molarity:

  • 200 mL⇒ 200/1000 = 0.200 L
  • 0.0653 mol Mn(ClO₄)₂ / 0.200 L = 0.3267 M

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