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.

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