Question

An aqueous solution of aluminum nitrite is prepared by dissolving 5.54 g of aluminum nitrite in 2.72×10² g of water. The density of the solution is 1.62 g mL⁻¹

An aqueous solution of lithium fluoride is prepared by dissolving 9.33 g of lithium fluoride in 5.39×10² g of water. The density of the solution is 1.34 g mL⁻¹.

Determine the molarity (in mol/L) of lithium fluoride in the solution.

Answer 1

The molarity of lithium fluoride in the solution is found to be 0.879 M, after calculating the number of moles of LiF and the volume of the solution in liters.

Step-by-step explanation:

To calculate the molarity of lithium fluoride in the solution, we need to use the formula:

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

Firstly, find the number of moles of lithium fluoride (LiF) by using its molar mass:

Moles of LiF = mass (g) / molar mass (g/mol)
Moles of LiF = 9.33 g / 25.94 g/mol = 0.3597 mol

To find the volume of the solution, we use the solution's mass and density:

Volume (mL) = mass of solution (g) / density (g/mL)
Mass of solution = mass of solute + mass of solvent = 9.33 g + 539 g = 548.33 g
Volume of solution = 548.33 g / 1.34 g/mL = 409.2 mL

Convert this volume into liters:

Volume (L) = 409.2 mL / 1000 mL/L = 0.4092 L

Now we can find the molarity:

M = 0.3597 mol / 0.4092 L = 0.879 M

Hence, the molarity of lithium fluoride in the solution is 0.879 M.