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Quantity of sugar from optical rotation

Short answer

Concentration C = θ/(l [α]) = 14.5°/(3 dm × 65°) = 0.0744 g/cm³. Mass of sugar in 50 cm³ is m = C V = 3.72 g.

A 30 cm long polarimeter tube containing 50 cm³ of sugar solution produces an optical rotation of 14.5° when placed on a polarimeter tube. If the specific rotation of sugar solution is 65°, calculate the quantity of sugar contained in the tube.

Specific rotation [α] = θ / (l C), where θ is the observed rotation (degrees), l is the tube length in dm and C the concentration in g/cm³.

Given: l = 30 cm = 3 dm, V = 50 cm³, θ = 14.5°, [α] = 65°.

  1. Concentration:
C = θ / (l [α]) = 14.5 / (3 × 65) = 14.5 / 195 = 0.07436 g/cm³
  1. Mass of sugar in the tube:
m = C V = 0.07436 × 50 = 3.72 g

Answer: the tube contains about 3.72 g of sugar.

✓ Answer checked by Claude Opus 5.5 (high effort)

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