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Strength of sugar solution

Short answer

Concentration c = θ/(l[α]) with l in dm. l = 2 dm, θ = 20°, [α] = 66°, so c = 20/(2×66) ≈ 0.15 g/cm³ (≈ 151.5 kg/m³).

A 20cm long tube having sugar solution rotates the plane of polarization by 20°. If specific rotation of sugar is 66°, calculate the strength of the solution.

Specific rotation [α] is the rotation produced by a 1 dm tube containing the solution at 1 g/cm³ concentration:

[α] = θ / (l × c)

so c = θ / (l × [α]), with l in decimetres and c in g/cm³.

Given: l = 20 cm = 2 dm, θ = 20°, [α] = 66° dm⁻¹ (g/cm³)⁻¹.

c = θ / (l [α])
  = 20 / (2 × 66)
  = 0.1515 g/cm³

In SI units this is 0.1515 × 1000 = 151.5 kg/m³.

Answer: strength (concentration) of the solution ≈ 0.15 g/cm³ (≈ 151.5 kg/m³, about 15 g per 100 cm³).

✓ Answer checked by Claude Opus 5.5 (high effort)

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