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

The molarity of the solution is 0.42 M. The molality of the solution is 0.46 m.

a)To obtain the molarity of the solution;

Number of moles = mass of glucose/molar mass of glucose = 28.8g/180 g/mol

= 0.16 moles

Volume of solution = 380mL or 0.38 L

Molarity = number of moles /volume =  0.16 moles/ 0.38 L  = 0.42 M

b) Molality of the solution;

Mass of solvent in Kg = 350g/1000 = 0.35 Kg

Molality = number of moles/mass of solution in kilogram = 0.16 moles/ 0.35 Kg = 0.46 m

c) percent by mass = mass of solute/mass of solution× 100

= 28.8g/(350g +28.8g)  × 100 = 7.6 %

d) Mole fraction

Number of moles of water = 350g/ 18 g/mol = 19.44 moles

Total number of moles = 19.44 moles + 0.16 moles = 19.6 moles

Mole fraction of glucose =  0.16 moles/19.6 moles = 0.0082

e) Mole percent

Mole fraction  × 100 =  0.0082  × 100 = 0.82%

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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 data:

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 %