Question

What volume in liters of a 0.320 M Mg(NO₃)₂ solution contains 45 g of Mg(NO₃)₂?

A) 70.31 L
B) 140.62 L
C) 35.16 L
D) 17.58 L

Answer 1

The volume of a 0.320 M Mg(NO₃)₂ solution containing 45 g of Mg(NO₃)₂ is approximately 0.947 liters.

Step-by-step explanation:

To find the volume of a 0.320 M Mg(NO₃)₂ solution that contains 45 g of Mg(NO₃)₂, we need to use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

First, we need to calculate the moles of Mg(NO₃)₂ by using its molar mass:

Molar Mass of Mg(NO₃)₂ = 24.31 g/mol (Mg) + 2(14.01 g/mol (N) + 3(16.00 g/mol (O)) = 148.31 g/mol

Moles of Mg(NO₃)₂ = mass (g) / molar mass (g/mol) = 45 g / 148.31 g/mol = 0.303 mol

Now, we can rearrange the formula to solve for the volume of the solution:

Volume of solution (L) = moles of solute / Molarity = 0.303 mol / 0.320 M = 0.9469 L

Therefore, the volume of a 0.320 M Mg(NO₃)₂ solution that contains 45 g of Mg(NO₃)₂ is approximately 0.947 liters.