A 2.40 M aqueous solution of LiCl has a density of 1.0538 g/mL. If the total mass of the solution is 54.0 g, what masses of solute and solvent are present?

5.21 g of solute (LiCl)

48.79 g of solvent (water)

Explanation:

This is our information

[LiCl] = 2.40 M → 2.40 moles of salt in 1L of solution

Density of solution: 1.0538 g/mL (solution mass / solution volume)

54 g → solution mass

Let's determine solution volume with density

1.0538 g/mL = solution mass / solution volume

1.0538 g/mL = 54 g / solution volume

Solution volume = 54 g / 1.0538 g/mL → 51.2 mL

Now, we can know the mass of solute, by molarity.

In 1 L of solution (1000 mL) we know that we have 2.40 mol of chloride.

Then, how many moles of chloride, do we have in 51.2mL of solution. We make a rule of three:

1000 mL has 2.40 moles of LiCl

51.2 mL would have (51.2. 2.40) / 1000 = 0.123 moles of solute

We apply molar mass to know the mass (mol. molar mass)

0.123 moles. 42.39 g/m = 5.21 g of LiCl

Finally solute mass + solvent mass = solution mass

5.21 g LiCl + solvent mass = 54 g

54 g - 5.21 g = solvent mass → 48.79 g