The mole fraction of a non electrolyte (MM 101.1 g/mol) in an aqueous solution is 0.0194. The solution's density is 1.0627 g/mL. Calculate the molarity of the solution.

Question

Chemistry

Ali Buchanan

30 April, 05:27

The mole fraction of a non electrolyte (MM 101.1 g/mol) in an aqueous solution is 0.0194. The solution's density is 1.0627 g/mL. Calculate the molarity of the solution.

+4

Answers (1)

Know the Answer?

Alma · Answer 1

Molarity for the solution is 1.05 mol/L

Explanation:

Mole fraction of solute = 0.0194

Solution's density = 1.0627 g/mL

We must know that sum of mole fraction = 1

Mole fraction of solute + Mole fraction of solvent = 1

0.0194 + Mole fraction of solvent = 1

Mole fraction of solvent = 1 - 0.0194 → 0.9806

Molarity is mol of solute in 1L of solution, so we have to determine solution's volume in L

With molar mass we can determine the mass of solute and solvent and then, the solution's mass

0.0194 mol. 101.1 g / mol = 1.96 g of non electrolyte solute

0.9806 mol. 18 g/mol = 17.65 g of water

Mass of solution = mass of solute + mass of solvent

1.96 g + 17.65 g = 19.6 g (mass of solution)

Solution's density = Solution's mass / Solution's volume

1.0627 g/mL = 19.6 g / Solution's volume

Solution's volume = 19.6 g / 1.0627 g/mol →18.4 mL

Let's convert the mass from mL to L

18.4mL. 1L / 1000 mL = 0.0184 L

We have the moles of solute, so let's determine molarity

mol/L → 0.0194 mol / 0.0184 L = 1.05 M