Answer :
To determine the boiling point of the solution, we need to look at how the presence of a solute (glucose in this case) affects the boiling point of the solvent (water). This is because adding a solute to a solvent typically raises its boiling point. This phenomenon is known as boiling point elevation.
Here’s a step-by-step breakdown of how you would calculate the boiling point of this glucose solution:
1. Determine the Molar Mass of Glucose (C₆H₁₂O₆):
- Carbon (C) has an atomic mass of 12, and there are 6 carbon atoms in glucose.
- Hydrogen (H) has an atomic mass of 1, and there are 12 hydrogen atoms in glucose.
- Oxygen (O) has an atomic mass of 16, and there are 6 oxygen atoms in glucose.
So, the molar mass of glucose is:
[tex]\[
(12 \times 6) + (1 \times 12) + (16 \times 6) = 180 \text{ g/mol}
\][/tex]
2. Calculate the Moles of Glucose:
To find the number of moles of glucose, divide the mass of glucose by its molar mass:
[tex]\[
\text{Moles of glucose} = \frac{70 \text{ g}}{180 \text{ g/mol}} \approx 0.3889 \text{ moles}
\][/tex]
3. Calculate the Molality of the Solution:
Molality is defined as the number of moles of solute per kilogram of solvent. You have 98.3 g of water, which is 0.0983 kg.
[tex]\[
\text{Molality} = \frac{0.3889 \text{ moles}}{0.0983 \text{ kg}} \approx 3.956 \text{ mol/kg}
\][/tex]
4. Find the Boiling Point Elevation:
The boiling point elevation ([tex]\(\Delta T_b\)[/tex]) can be calculated using the formula:
[tex]\[
\Delta T_b = K_b \times \text{molality}
\][/tex]
where [tex]\(K_b\)[/tex] is the ebullioscopic constant for water, which is 0.512 °C·kg/mol.
[tex]\[
\Delta T_b = 0.512 \times 3.956 \approx 2.026 \text{ °C}
\][/tex]
5. Calculate the New Boiling Point of the Solution:
The boiling point of pure water is 100 °C. So, by adding the boiling point elevation to the boiling point of pure water, we get the new boiling point of the solution:
[tex]\[
\text{Boiling point of the solution} = 100 + 2.026 \approx 102.026 \text{ °C}
\][/tex]
Therefore, the boiling point of the solution is approximately 102.03 °C.
Here’s a step-by-step breakdown of how you would calculate the boiling point of this glucose solution:
1. Determine the Molar Mass of Glucose (C₆H₁₂O₆):
- Carbon (C) has an atomic mass of 12, and there are 6 carbon atoms in glucose.
- Hydrogen (H) has an atomic mass of 1, and there are 12 hydrogen atoms in glucose.
- Oxygen (O) has an atomic mass of 16, and there are 6 oxygen atoms in glucose.
So, the molar mass of glucose is:
[tex]\[
(12 \times 6) + (1 \times 12) + (16 \times 6) = 180 \text{ g/mol}
\][/tex]
2. Calculate the Moles of Glucose:
To find the number of moles of glucose, divide the mass of glucose by its molar mass:
[tex]\[
\text{Moles of glucose} = \frac{70 \text{ g}}{180 \text{ g/mol}} \approx 0.3889 \text{ moles}
\][/tex]
3. Calculate the Molality of the Solution:
Molality is defined as the number of moles of solute per kilogram of solvent. You have 98.3 g of water, which is 0.0983 kg.
[tex]\[
\text{Molality} = \frac{0.3889 \text{ moles}}{0.0983 \text{ kg}} \approx 3.956 \text{ mol/kg}
\][/tex]
4. Find the Boiling Point Elevation:
The boiling point elevation ([tex]\(\Delta T_b\)[/tex]) can be calculated using the formula:
[tex]\[
\Delta T_b = K_b \times \text{molality}
\][/tex]
where [tex]\(K_b\)[/tex] is the ebullioscopic constant for water, which is 0.512 °C·kg/mol.
[tex]\[
\Delta T_b = 0.512 \times 3.956 \approx 2.026 \text{ °C}
\][/tex]
5. Calculate the New Boiling Point of the Solution:
The boiling point of pure water is 100 °C. So, by adding the boiling point elevation to the boiling point of pure water, we get the new boiling point of the solution:
[tex]\[
\text{Boiling point of the solution} = 100 + 2.026 \approx 102.026 \text{ °C}
\][/tex]
Therefore, the boiling point of the solution is approximately 102.03 °C.
