- 2068 Shrawan · 5 marks
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)
Related questions
- Quantity of sugar in 48 cm³ solution · Engineering Physics, chapter 4 · asked 3×
- Quantity of sugar from optical rotation · Engineering Physics, chapter 4 · asked 3×
- Shear strength on horizontal plane, WT 3.5 m · Soil Mechanics, chapter 10
- Settling tank 20 m long, 3.5 m deep; 0.02 and 0.01 mm · Water Supply Engineering, chapter 6 · asked 3×
- Stresses on 20 degree plane; principal and max shear · Strength of Materials, chapter 4 · asked 4×
- 3D rotation about axis (1,2,3)-(10,20,30) · Computer Graphics, chapter 4