To solve this problem, we need to use the formula for molarity which is moles of solute/volume of solution in liters.

Step 1: Convert the mass of the solution to volume. Assuming the density of the solution is approximately 1 g/mL (which is the density of water and a common approximation unless otherwise specified), 603 g is approximately 603 mL. Convert this to liters by dividing by 1000, which gives 0.603 L.

Step 2: Calculate the moles of Mn(ClO₄)₂ in the solution. We can do this by multiplying the molarity of the solution by the volume in liters. 1.55 M * 0.603 L = 0.93465 moles of Mn(ClO₄)₂.

Step 3: Convert moles of Mn(ClO₄)₂ to grams. The molar mass of Mn(ClO₄)₂ is approximately 271.3 g/mol. Multiply the number of moles by the molar mass to get the mass in grams. 0.93465 moles * 271.3 g/mol = approximately 253.5 g.

So, there are approximately 253.5 grams of Mn(ClO₄)₂ in 603 g of 1.55 M solution.

Question

To solve this problem, we need to use the formula for molarity which is moles of solute/volume of solution in liters.

Step 1: Convert the mass of the solution to volume. Assuming the density of the solution is approximately 1 g/mL (which is the density of water and a common approximation unless otherwise specified), 603 g is approximately 603 mL. Convert this to liters by dividing by 1000, which gives 0.603 L.

Step 2: Calculate the moles of Mn(ClO₄)₂ in the solution. We can do this by multiplying the molarity of the solution by the volume in liters. 1.55 M * 0.603 L = 0.93465 moles of Mn(ClO₄)₂.

Step 3: Convert moles of Mn(ClO₄)₂ to grams. The molar mass of Mn(ClO₄)₂ is approximately 271.3 g/mol. Multiply the number of moles by the molar mass to get the mass in grams. 0.93465 moles * 271.3 g/mol = approximately 253.5 g.

So, there are approximately 253.5 grams of Mn(ClO₄)₂ in 603 g of 1.55 M solution.

Knowee AI · Accepted Answer

To solve this problem, we need to use the formula for molarity which is moles of solute/volume of solution in liters.

Step 1: Convert the mass of the solution to volume. Assuming the density of the solution is approximately 1 g/mL (which is the density of water and a common approximation unless otherwise specified), 603 g is approximately 603 mL. Convert this to liters by dividing by 1000, which gives 0.603 L.

Step 2: Calculate the moles of Mn(ClO₄)₂ in the solution. We can do this by multiplying the molarity of the solution by the volume in liters. 1.55 M * 0.603 L = 0.93465 moles of Mn(ClO₄)₂.

Step 3: Convert moles of Mn(ClO₄)₂ to grams. The molar mass of Mn(ClO₄)₂ is approximately 271.3 g/mol. Multiply the number of moles by the molar mass to get the mass in grams. 0.93465 moles * 271.3 g/mol = approximately 253.5 g.

So, there are approximately 253.5 grams of Mn(ClO₄)₂ in 603 g of 1.55 M solution.

How many grams of Mn(ClO₄)₂ are in 603 g of 1.55 M solution?

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