Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To determine the molality of a [tex]$D$[/tex]-glucose solution prepared by dissolving [tex]\(54.0\)[/tex] grams of [tex]\(D\)[/tex]-glucose ([tex]\(C_6H_{12}O_6\)[/tex]) in [tex]\(125.9\)[/tex] grams of water, follow these detailed steps:
1. Determine the molar mass of [tex]\(D\)[/tex]-glucose ([tex]\(C_6H_{12}O_6\)[/tex]):
The molar mass is given as [tex]\(180.16\)[/tex] g/mol.
2. Calculate the number of moles of [tex]\(D\)[/tex]-glucose:
Use the formula:
[tex]\[ \text{moles of \(D\)-glucose} = \frac{\text{mass of \(D\)-glucose}}{\text{molar mass of \(D\)-glucose}} \][/tex]
Given the mass of [tex]\(D\)[/tex]-glucose ([tex]\(54.0\)[/tex] grams):
[tex]\[ \text{moles of \(D\)-glucose} = \frac{54.0 \, \text{g}}{180.16 \, \text{g/mol}} \approx 0.2997 \, \text{mol} \][/tex]
3. Convert the mass of the solvent (water) to kilograms:
Given the mass of the solvent ([tex]\(125.9\)[/tex] grams):
[tex]\[ \text{solvent mass} = \frac{125.9 \, \text{g}}{1000} = 0.1259 \, \text{kg} \][/tex]
4. Calculate the molality ([tex]\(m\)[/tex]) of the solution:
Molality is defined as the number of moles of solute per kilogram of solvent. Use the formula:
[tex]\[ \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \][/tex]
Substitute the calculated values:
[tex]\[ \text{molality} = \frac{0.2997 \, \text{mol}}{0.1259 \, \text{kg}} \approx 2.38 \, m \][/tex]
Therefore, the molality of the [tex]\(D\)[/tex]-glucose solution is [tex]\(2.38 \, m\)[/tex]. The correct answer is:
[tex]\[ \boxed{2.38 \, m} \][/tex]
1. Determine the molar mass of [tex]\(D\)[/tex]-glucose ([tex]\(C_6H_{12}O_6\)[/tex]):
The molar mass is given as [tex]\(180.16\)[/tex] g/mol.
2. Calculate the number of moles of [tex]\(D\)[/tex]-glucose:
Use the formula:
[tex]\[ \text{moles of \(D\)-glucose} = \frac{\text{mass of \(D\)-glucose}}{\text{molar mass of \(D\)-glucose}} \][/tex]
Given the mass of [tex]\(D\)[/tex]-glucose ([tex]\(54.0\)[/tex] grams):
[tex]\[ \text{moles of \(D\)-glucose} = \frac{54.0 \, \text{g}}{180.16 \, \text{g/mol}} \approx 0.2997 \, \text{mol} \][/tex]
3. Convert the mass of the solvent (water) to kilograms:
Given the mass of the solvent ([tex]\(125.9\)[/tex] grams):
[tex]\[ \text{solvent mass} = \frac{125.9 \, \text{g}}{1000} = 0.1259 \, \text{kg} \][/tex]
4. Calculate the molality ([tex]\(m\)[/tex]) of the solution:
Molality is defined as the number of moles of solute per kilogram of solvent. Use the formula:
[tex]\[ \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \][/tex]
Substitute the calculated values:
[tex]\[ \text{molality} = \frac{0.2997 \, \text{mol}}{0.1259 \, \text{kg}} \approx 2.38 \, m \][/tex]
Therefore, the molality of the [tex]\(D\)[/tex]-glucose solution is [tex]\(2.38 \, m\)[/tex]. The correct answer is:
[tex]\[ \boxed{2.38 \, m} \][/tex]
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.