---
title: "Quantity of sugar from optical rotation"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:sugar-rotation-quantity-thirty-cm"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/sugar-rotation-quantity-thirty-cm
site: BCT Engineering
---

# Quantity of sugar from optical rotation

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2074 Chaitra, 5 marks
- Video: [Specific Rotation and Observed Rotation Calculations in Optical Activity](https://www.youtube.com/watch?v=jzdYqLO9OY8) — Leah4sci
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/sugar-rotation-quantity-thirty-cm
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

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.

## 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.

## Answer

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:

```text
C = θ / (l [α]) = 14.5 / (3 × 65) = 14.5 / 195 = 0.07436 g/cm³
```

2. Mass of sugar in the tube:

```text
m = C V = 0.07436 × 50 = 3.72 g
```

**Answer: the tube contains about 3.72 g of sugar.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula relates rotation θ, length l, concentration C and specific rotation [α]?
   A. [α] = θ / (l C)
   B. [α] = θ l C
   C. [α] = l C / θ
   D. [α] = θ C / l
   Answer: A. Specific rotation = rotation per dm per g/cm^3. (Key point: Formula)

2. What is the 30 cm tube length in dm?
   A. 30 dm
   B. 0.3 dm
   C. 3 dm
   D. 300 dm
   Answer: C. 10 cm = 1 dm, so 30 cm = 3 dm. (Key point: Unit conversion)

3. What is C for θ = 14.5°, l = 3 dm, [α] = 65°?
   A. 0.0744 g/cm^3
   B. 0.223 g/cm^3
   C. 13.4 g/cm^3
   D. 0.0248 g/cm^3
   Answer: A. C = 14.5/(3×65) = 0.07436 g/cm^3. (Key point: Concentration)

4. How much sugar is in 50 cm^3?
   A. 0.0015 g
   B. 3.72 g
   C. 37.2 g
   D. 671 g
   Answer: B. m = C × V = 0.07436 × 50 = 3.72 g. (Key point: Mass)

## Close variants set in other exams

- [Quantity of sugar in 48 cm³ solution](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asugar-rotation-quantity-two-hundred-mm) — Engineering Physics, chapter 4, asked 3×

## Related in Engineering Physics

- [Strength of sugar solution](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asugar-strength-twenty-cm-tube) — Engineering Physics, chapter 4