A solution is prepared by dissolving 28.8g of glucose (C6H12O6) in 350g of water. The final volume of the solution is 380mL. For this solution, calculate each of the following: molarity, Molality, percent by mass, mole fraction, mole percent

Molarity → 0.410 M

Molality → 0.44 m

Percent by mass → 7.61 g

Mole fraction (Xm) = 7.96*10⁻³

Mole percent = 0.79 %

Explanation:

We analyse dа ta:

28.8 g of glucose → To determine molarity, molality, mole fraction, mole percent we need to find out the moles:

28.2 g. 1mol / 180 g = 0.156 moles of glucose

350 g of water → Mass of solvent.

We convert from g to kg in order to determine molality = 350 g / 1000 = 0.350 kg

We also need the moles of solvent: 350 g / 18 g/mol = 19.44 moles

380 mL of solution → Volume of solution; to determine the molarity we need the volume in L → 380 mL / 1000 = 0.380L

Solution mass = Solute mass + Solvent mass

28.2 g + 350 g = 378.2 g

Total moles = Moles of solute + Moles of solvent

0.156 + 19.44 = 19.596 moles

Molarity → Moles of solute in 1L of solution → 0.156 mol / 0.380L = 0.410 M

Molality → Moles of solute in 1kg of solvent → 0.156 mol / 0.350 kg = 0.44 m

Percent by mass → Mass of solute in 100 g of solution

(28.8g / 378.2g). 100 = 7.61 g

Mole fraction (Xm) = Moles of solute / Total moles → 0.156 mol / 19.596 moles = 7.96*10⁻³

Mole percent = Xm. 100 → 7.96*10⁻³. 100 = 0.79 %