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What is the %(w/v) of a 0.500 M glucose (C6H12O6) solution? MM of glucose is 180 g/mol. 18.0% 1.80% 9.00% 90.0%

What is the %(w/v) of a 0.500 M glucose (C6H12O6) solution? MM of glucose is 180 g/mol. 18.0% 1.80% 9.00% 90.0%

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Question:** What is the % (w/v) of a 0.500 M glucose (C₆H₁₂O₆) solution? The molar mass (MM) of glucose is 180 g/mol. **Options:** - ○ 18.0% - ○ 1.80% - ○ 9.00% - ○ 90.0% **Explanation:** To solve this problem, follow these steps: 1. Calculate the mass of glucose in grams in one liter of the solution: \[ \text{Mass} = \text{Molarity} \times \text{Molar Mass} \times \text{Volume} \] Given: Molarity (M) = 0.500 M and Molar Mass (MM) of glucose = 180 g/mol For 1 liter of solution (Volume = 1 L): \[ \text{Mass} = 0.500 \, \text{M} \times 180 \, \text{g/mol} \times 1 \, \text{L} = 90 \, \text{g} \] 2. The % (w/v) concentration is defined as grams of solute (glucose) per 100 mL of solution. \[ \%\ (w/v) = \left( \frac{\text{Mass of solute}}{\text{Volume of solution in mL}} \right) \times 100 \] \[ \%\ (w/v) = \left( \frac{90 \, \text{g}}{1000 \, \text{mL}} \right) \times 100 \] \[ \%\ (w/v) = 9\% \] Thus, the correct answer is: - ○ 9.00%
Transcribed Image Text:**Question:** What is the % (w/v) of a 0.500 M glucose (C₆H₁₂O₆) solution? The molar mass (MM) of glucose is 180 g/mol. **Options:** - ○ 18.0% - ○ 1.80% - ○ 9.00% - ○ 90.0% **Explanation:** To solve this problem, follow these steps: 1. Calculate the mass of glucose in grams in one liter of the solution: \[ \text{Mass} = \text{Molarity} \times \text{Molar Mass} \times \text{Volume} \] Given: Molarity (M) = 0.500 M and Molar Mass (MM) of glucose = 180 g/mol For 1 liter of solution (Volume = 1 L): \[ \text{Mass} = 0.500 \, \text{M} \times 180 \, \text{g/mol} \times 1 \, \text{L} = 90 \, \text{g} \] 2. The % (w/v) concentration is defined as grams of solute (glucose) per 100 mL of solution. \[ \%\ (w/v) = \left( \frac{\text{Mass of solute}}{\text{Volume of solution in mL}} \right) \times 100 \] \[ \%\ (w/v) = \left( \frac{90 \, \text{g}}{1000 \, \text{mL}} \right) \times 100 \] \[ \%\ (w/v) = 9\% \] Thus, the correct answer is: - ○ 9.00%
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