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6 September, 17:06

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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  1. 6 September, 17:34
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    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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