Question

Maximum solubility of sugar at $20^\circ C$ is equal to……..

• A. 43.2%
• B. 82%
• C. 67.5%
• D. 75.6%

Hint:

Remember the "2 to 1" rule for sugar: At room temp, you can dissolve roughly 2 parts sugar in 1 part water. $(2/3) \times 100 \approx 67\%$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the weight-by-weight percentage of sucrose in a saturated solution at a standard room temperature of $20^\circ C$.

Step 2: Detailed Explanation:

• Sucrose Solubility: Sucrose is highly soluble in water, and its solubility increases significantly with temperature.

• Data at $20^\circ C$: At $20^\circ C$, approximately 200 grams of sucrose can be dissolved in 100 grams of water.

• Percentage Calculation:
- Mass of solute (sugar) = 200 g.
- Mass of solvent (water) = 100 g.
- Total mass of solution = 300 g.
- Percentage ($\% w/w$) = $\frac{\text{Mass of Solute}}{\text{Total Mass}} \times 100$
- Percentage = $\frac{200}{300} \times 100 = 66.66\%$.

• Choice Analysis: Standard physical chemistry tables list the solubility of sucrose at $20^\circ C$ as 67.5% (w/w). This corresponds to Option (C).

• Industrial Importance: This value defines the limit for concentrated syrups. Exceeding this concentration (supersaturation) leads to crystallization, unless temperature is increased.

Step 3: Final Answer:
The maximum solubility of sucrose in water at $20^\circ C$ is approximately 67.5% by weight.