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%

Learn more: link

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 %

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.