Question

A solution of fructose, C6H12O6, a sugar found in many fruits, is made by dissolving 34.0 g of fructose in 1.00 kg of water. What is the molal concentration of fructose the solution? What is the mole fraction of fructose in the solution? What is the mass percent of fructose in the solution? What is the molarity of the fructose solution if the density of the solution is 1.0078 g/mL?

Answer 1

Answer:

Molality → 0.188 m

Mole fraction of fructose → 0.00337

Mass percent of fructose in solution → 3.29 %

Molarity → 0.183 M

Explanation:

Solute → 34 g of fructose

Solvent → 1000 g of water

Solution → 1000 g of water + 34 g of fructose = 1034 g of solution.

We take account density to calculate, the solution's density

1.0078 g/mL = 1034 g / mL

1034 g / 1.0078 g/mL = 1026 mL

Molal concentration → moles of solute in 1kg of solvent

Moles of fructose = mass of fructose / molar mass

34 g/ 180g/mol = 0.188 mol

0.188 mol/1kg = 0.188 m

Mole fraction of fructose = Moles of fructose / Total moles

We determine the moles of water

Moles of water = 1000 g / 18 g = 55.5 mol

Total moles = moles of fructose + moles of water

0.188 mol + 55.5 mol = 55.743 mol

0.188 mol / 55.743 mol = 0.00337

Mass percent = mass of fructose in 100 g of solution

(Mass of fructose / Total mass ) . 100 = (34 g /1034 g) . 100 = 3.29 %

Molarity = Moles of solute in 1L of solution

We can also say mmol of solute in 1 mL of solution

0.188 mol of fructose = 188 mmol of fructose

Molarity = 188 mmol / 1026 mL of solution = 0.183 M