100gm of an aq. solution of sugar contains 50% sugar by mass. How much water should be evaporated get 60% sugar solution by mass?

Options

A. 10g
B. 20g
C. 30g
D. None of these

Explanation

To solve the problem of how much water should be evaporated to obtain a 60% sugar solution from a 50% sugar solution, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the initial composition of the solution:**
- We have 100 grams of an aqueous sugar solution that contains 50% sugar by mass.
- Therefore, the mass of sugar in the solution is:
\[ \text{Mass of sugar} = 50\% \text{ of } 100 \text{ g} = 50 \text{ g} \]
- The mass of water in the solution is:
\[ \text{Mass of water} = 100 \text{ g} - 50 \text{ g} = 50 \text{ g} \]

2. **Determine the final composition required:**
- We want to achieve a solution that is 60% sugar by mass.
- This means that in the final solution, the mass of sugar should be 60% of the total mass of the solution.
- Let \( x \) be the total mass of the final solution. Therefore, we have:
\[ \text{Mass of sugar} = 60\% \text{ of } x = 0.6x \]
- Since the mass of sugar does not change during evaporation, we still have 50 g of sugar:
\[ 0.6x = 50 \text{ g} \]

3. **Calculate the total mass of the final solution:**
- Rearranging the equation gives:
\[ x = \frac{50 \text{ g}}{0.6} = \frac{50}{0.6} \approx 83.33 \text{ g} \]

4. **Determine the mass of water in the final solution:**
- The mass of water in the final solution can be calculated as:
\[ \text{Mass of water} = x - \text{Mass of sugar} = 83.33 \text{ g} - 50 \text{ g} = 33.33 \text{ g} \]

5. **Calculate the amount of water to be evaporated:**
- Initially, we had 50 g of water, and now we need 33.33 g of water in the final solution.
- The amount of water that should be evaporated is:
\[ \text{Water evaporated} = \text{Initial mass of water} - \text{Final mass of water} = 50 \text{ g} - 33.33 \text{ g} \approx 16.67 \text{ g} \]

### Final Answer:
Approximately 16.67 grams of water should be evaporated to obtain a 60% sugar solution by mass.