Question

How many grams of Mg(NO3), are necessary to
make 1500 mL of a 0.50 M solution?

Answer 1

1.1 × 10² g

Step-by-step explanation:

Step 1: Given data

• Concentration of the solution (C): 0.50 M (0.50 mol/L)

• Volume of solution (V): 1500 mL (1.500 L)

Step 2: Calculate the moles of Mg(NO₃)₂ (solute)

Molarity is equal to the moles of solute (n) divided by the liters of solution.

C = n/V

n = C × V

n = 0.50 mol/L × 1.500 L = 0.75 mol

Step 3: Calculate the mass corresponding to 0.75 moles of Mg(NO₃)₂

The molar mass of Mg(NO₃)₂ is 148.3 g/mol.

0.75 mol × 148.3 g/mol = 1.1 × 10² g