Determining the Mass of Mg(NO3)2 Present in a Solution

What is the mass of Mg(NO3)2 present in 129 mL of a 0.450 M solution of Mg(NO3)2?

A. 42.5 g

B. 0.0581 g

C. 5.01 g

D. 8.61 g

Answer:

The correct answer is D. 8.61 g.

To determine the mass of Mg(NO3)2 present in the solution, we need to use the formula: Mass = Volume × Concentration × Molar mass. The mass of Mg(NO3)2 present in 129 mL of a 0.450 M solution is approximately 8.613 grams.

Given: Volume of solution = 129 mL. Concentration of Mg(NO3)2 solution = 0.450 M. First, we need to convert the volume of the solution from milliliters (mL) to liters (L): 129 mL = 129/1000 L = 0.129 L. Next, we need to calculate the molar mass of Mg(NO3)2: Molar mass of Mg(NO3)2 = (1 × atomic mass of Mg) + (2 × atomic mass of N) + (6 × atomic mass of O) Molar mass of Mg(NO3)2 = (1 × 24.31 g/mol) + (2 × 14.01 g/mol) + (6 × 16.00 g/mol) Molar mass of Mg(NO3)2 = 148.31 g/mol Now, we can calculate the mass of Mg(NO3)2: Mass = Volume × Concentration × Molar mass Mass = 0.129 L × 0.450 M × 148.31 g/mol Mass = 8.613 g. Therefore, the mass of Mg(NO3)2 present in 129 mL of a 0.450 M solution is approximately 8.613 grams.