How many milliliters of an aqueous solution of 0.206 M manganese(II) sulfate is needed to obtain 2.49 grams of the salt?     I would like to see this worked out please.

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To solve this problem, we can use the formula in getting the molarity of a solution.

`molarity = (mol es)/(L I t e r s o f s o l u t I o n)`

Since we are looking for the volume of the solution in milliliters, we can transform...

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To solve this problem, we can use the formula in getting the molarity of a solution.

`molarity = (mol es)/(L I t e r s o f s o l u t I o n)`

Since we are looking for the volume of the solution in milliliters, we can transform the formula into:

`Liters of solution = (mol es)/(molarity)`

We also know that the moles can be obtained from:

`mol e s = (grams)/(molar mass)`

Given:

Molarity (M) = 0.206 M

Mass of Manganese (II) Sulfate = 2.49

Molar mass of Manganese (II) Sulfate = 151 g/mol (from the periodic table)

First we can get the number of moles of the compound:

Moles MnSO4 = 2.49/151 = 0.01649 moles MnSO4

Then we can now get the volume of the solution needed:

`Liters of solution = (mol es)/(molarity)`

`Liters of solution = (0.01649)/(0.206)`

`Liters of solution = 0.080 Liters`

 

Since the question is asking for the milliliters of the solution, we can just multiply the answer by 1000.

1L = 1000 mL

Answer: 80.0 Milliliters of solution

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