To find the boiling point of the solution, we need to calculate how much the boiling point increases when the glucose is added. This is called "boiling point elevation."
Here is the step-by-step solution:
Step 1: Find the molar mass of Glucose ($C_6H_{12}O_6$).
First, we determine the mass of one mole of glucose by adding up the atomic masses from the periodic table:
* Carbon (C): $12.01 \text{ g/mol} \times 6 = 72.06 \text{ g/mol}$
* Hydrogen (H): $1.008 \text{ g/mol} \times 12 = 12.096 \text{ g/mol}$
* Oxygen (O): $16.00 \text{ g/mol} \times 6 = 96.00 \text{ g/mol}$
Total Molar Mass = $72.06 + 12.096 + 96.00 \approx 180.16 \text{ g/mol}$.
Step 2: Calculate the number of moles of Glucose.
Use the formula: $\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$
$$ \text{Moles} = \frac{45.0 \text{ g}}{180.16 \text{ g/mol}} \approx 0.2498 \text{ mol} $$
Step 3: Calculate the Molality ($m$) of the solution.
Molality is defined as moles of solute per kilogram of solvent.
* Mass of water (solvent) = $250.0 \text{ g} = 0.2500 \text{ kg}$
$$ m = \frac{0.2498 \text{ mol}}{0.2500 \text{ kg}} \approx 0.9992 \text{ mol/kg} $$
Step 4: Calculate the Boiling Point Elevation ($\Delta T_b$).
We use the formula found in the worksheet's table for boiling point elevation:
$$ \Delta T_b = K_b \cdot m $$
* From the table, the constant for water ($K_b$) is $0.51^\circ\text{C}/m$.
* Since glucose is a covalent compound (non-electrolyte), it does not break into ions, so we do not need to adjust for the van 't Hoff factor ($i=1$).
$$ \Delta T_b = 0.51^\circ\text{C}/m \times 0.9992 \text{ m} \approx 0.51^\circ\text{C} $$
Step 5: Determine the final Boiling Point.
The normal boiling point of pure water is $100.00^\circ\text{C}$. We add the elevation amount to this temperature.
$$ \text{New Boiling Point} = 100.00^\circ\text{C} + 0.51^\circ\text{C} = 100.51^\circ\text{C} $$
Final Answer:
100.51°C
Parent Tip: Review the logic above to help your child master the concept of colligative properties worksheet.
