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Calculate M 0.530 g of Mg(NO3)2 in 250.0 mL of solution

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User Krassi
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1 Answer

6 votes

Answer:

To calculate the molarity (M) of a solution, you need to know the amount of solute (in moles) and the volume of the solution (in liters).

Given:

Mass of Mg(NO3)2 = 0.530 g

Volume of solution = 250.0 mL = 0.250 L

To calculate the number of moles of Mg(NO3)2, you need to convert the mass to moles using the molar mass of Mg(NO3)2.

Molar mass of Mg(NO3)2:

Mg: 24.31 g/mol

N: 14.01 g/mol

O: 16.00 g/mol (x3 for three oxygen atoms)

2 nitrogen atoms + 6 oxygen atoms = (14.01 g/mol x 2) + (16.00 g/mol x 6) = 46.02 g/mol

Now, calculate the moles of Mg(NO3)2:

Moles = Mass / Molar mass

Moles = 0.530 g / 46.02 g/mol

Now we have the moles of Mg(NO3)2, and we can calculate the molarity (M) using the formula:

Molarity (M) = Moles / Volume

Molarity = (0.530 g / 46.02 g/mol) / 0.250 L

Calculate the value:

Molarity = 0.0115 mol / 0.250 L

Molarity = 0.046 M (rounded to three decimal places)

Therefore, the molarity of the solution containing 0.530 g of Mg(NO3)2 in 250.0 mL of solution is approximately 0.046 M.

answered
User Ron Harlev
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