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Quantity of sugar in 48 cm³ solution

Short answer

Concentration C = θ / (l [α]) = 11° / (2 dm × 66° dm⁻¹(g/cm³)⁻¹) = 0.0833 g/cm³. Mass of sugar = C × volume = 0.0833 g/cm³ × 48 cm³ = 4.0 g.

A 200 mm long tube containing 48 cm³ of sugar solution produces an optical rotation of 11° when placed in a saccharimeter. If the specific rotation of sugar solution is 66°, calculate the quantity of sugar contained in the tube in the form of a solution.

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

Given: l = 200 mm = 2 dm, θ = 11°, [α] = 66° dm⁻¹ (g/cm³)⁻¹, volume V = 48 cm³.

  1. Concentration:
C = θ / (l [α]) = 11 / (2 × 66) = 0.0833 g/cm³
  1. Quantity (mass) of sugar in the tube:
m = C × V = 0.0833 × 48 = 4.0 g

Answer: the tube contains 4.0 g of sugar.

✓ Answer checked by Claude Opus 5.5 (high effort)

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