Question

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.

Answer 1

The molarity of the solution when 23.640 g of Mn(ClO4)2 · 6 H2O is added to 200.0 mL of water is 0.3014 M, calculated by dividing the moles of solute by the volume of solution in liters.

Step-by-step explanation:

To calculate the molarity of a solution when a certain amount of solute is added to a known volume of water, you must first determine the number of moles of solute. For the molarity of the resulting solution when 23.640 g of Mn(ClO4)2 · 6 H2O (molar mass approximately 392.13 g/mol) is dissolved in 200.0 mL of water, the calculation is as follows:

1. Calculate the moles of Mn(ClO4)2 · 6 H2O using its molar mass.
2. Divide the moles by the volume of the solution in liters to find the molarity.

To calculate the moles of Mn(ClO4)2 · 6 H2O: 23.640 g / 392.13 g/mol ≈ 0.06028 mol.

To find molarity (M), which is moles of solute per liter of solution: 0.06028 mol / 0.200 L = 0.3014 M.

Answer 2

0.3267 M

Step-by-step 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