Question

6 g of urea (molar mass = 60 g mol$^{-1}$) and 9 g of glucose (molar mass = 180 g mol$^{-1}$) were dissolved in 35 g of water. The mass percent of urea and glucose is respectively.

• A. 18, 12
• B. 6, 9
• C. 12, 18
• D. 9, 6

Hint:

To calculate mass percent, remember to use the formula: \(\text{Mass percent} = \frac{\text{Mass of substance}}{\text{Total mass of solution}} \times 100\).

Answer 1

The Correct Option is C

The mass percent of a substance is given by: \[ \text{Mass percent of substance} = \frac{\text{Mass of substance}}{\text{Mass of solution}} \times 100 \] - For urea: \[ \text{Mass percent of urea} = \frac{6}{6 + 9 + 35} \times 100 = \frac{6}{50} \times 100 = 12% \] - For glucose: \[ \text{Mass percent of glucose} = \frac{9}{6 + 9 + 35} \times 100 = \frac{9}{50} \times 100 = 18% \] Thus, the mass percent of urea is 12% and glucose is 18%.

Answer 2

Step 1: Given Data
- Mass of urea = 6 g
- Molar mass of urea = 60 g/mol
- Mass of glucose = 9 g
- Molar mass of glucose = 180 g/mol
- Mass of water = 35 g

Step 2: Calculate Total Mass of Solution
\[ \text{Total mass} = \text{mass of urea} + \text{mass of glucose} + \text{mass of water} = 6 + 9 + 35 = 50 \, \text{g} \]

Step 3: Calculate Mass Percent of Urea
\[ \text{Mass \% of urea} = \left( \frac{\text{mass of urea}}{\text{total mass}} \right) \times 100 = \frac{6}{50} \times 100 = 12\% \]

Step 4: Calculate Mass Percent of Glucose
\[ \text{Mass \% of glucose} = \left( \frac{\text{mass of glucose}}{\text{total mass}} \right) \times 100 = \frac{9}{50} \times 100 = 18\% \]

Step 5: Conclusion
The mass percent of urea and glucose in the solution are 12% and 18%, respectively.