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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(ClOâ‚„)â‚‚ 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(ClOâ‚„)₂·6Hâ‚‚O ÷ 361.84 g/mol = 0.0653 mol Mn(ClOâ‚„)₂·6Hâ‚‚O = mol Mn(ClOâ‚„)â‚‚

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

  • 200 mL⇒ 200/1000 = 0.200 L
  • 0.0653 mol Mn(ClOâ‚„)â‚‚ / 0.200 L = 0.3267 M

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