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a) An aqueous solution containing 10g of optically pure fructose was diluted to 500mL with water and placed in a polarimeter tube 20 cm long.The measured rotation was -5.20 degrees.Calculate the specific rotation of fructose.b) If this solution were mixed with 500 mL of a solution containing 5 g of racemic fructose,what would be the specific rotation of the resulting fructose mixture?What would be its optical purity?

a) Specific rotation of pure fructose = - 130°

b) Specific rotation of racemic fructose = 65°

Specific rotation of the racemic mixture = -65°

Optical purity = 50%

Explanation:

specific rotation = observed rotation/ (tube length in decimeters x concentration)

[α] = α/lc

Observed rotation, α = -5.2°

Tube length in decimeters, l = 2 decimeters (1 decimeter = 10cm)

Concentration in g/mL or g/cm³ = 10/500 = 0.02 g/mL

Specific rotation, [α] = -5.2/(2×0.02) = - 130°

b) The racemic fructose would rotate light in the opposite direction. If the two enantiomers have the same concentration, specific rotation of the racemic fructose would be -(-130°) = 130°, but the concentrations are different.

Concentration in g/mL or g/cm³ = 5/500 = 0.01 g/mL

At half of the concentration of optically pure fructose, specific rotation would be 0.5 × 130° = 65°

Total specific rotation of mixture = - 130° + 65° = -65°

Optical purity = (specific rotation of mixture/specific rotation of pure sample) × 100% = [-65/(-130)]× 100 = 50%

Hope this Helps!!!

Boiling T° of solution = 135.6°C

Explanation:

Formula for elevation boiling point is:

ΔT = Kb . m . i

ΔT =  Boiling point of solution - Boiling point of pure solvent

Kb = Boiling point elevation constant

m = molality →  moles of solute in 1kg of solvent

i = numbers of ions dissolved

FeCl₃ →  Fe³⁺  +  3Cl⁻

In the dissociation of the ionic salt, we determined 4 moles of ions dissolved.

3 for chlorides and 1 for iron. Then i = 4

m → We convert the mass of solute to moles:

90 g . 1mol / 162.2g = 0.555 moles

650 g of solvent = 0.650 kg of solvent

m = 0.555 mol/0.650kg → 0.85

We replace data at formula

Boiling T° of solution - 129.90°C = 1.67°C . kg/mol  .  0.85 mol/kg . 4

Boiling T° of solution = 1.67°C . kg/mol  .  0.85 mol/kg . 4 + 129.90°C

Boiling T° of solution = 135.6°C