Answer :
Great! Let's go through the steps to solve each question using the given data.
Table 6.1
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text {Solution} & \text {Lowest Temperature (}^{\circ} \mathrm{C}) & \text {1st Trial (}^{\circ} \mathrm{C}) & \text {2nd Trial (}^{\circ} \mathrm{C}) & \text {Average (}^{\circ} \mathrm{C}) & \Delta T_1 (^{\circ} \mathrm{C}) \\ \hline \text {Sucrose} & -7.9 & -1.6 & -1.6 & -1.6 & -7.9 - (-1.6) = -6.3 \\ \hline \mathrm{H}_2 \mathrm{O} & -8.0 & 0.0 & 0.0 & 0.0 & -8.0 - 0.0 = -8.0 \\ \hline \end{array} \][/tex]
### Question 1:
Was there supercooling? Would you expect the water or the sugar solution to have the most supercooling? Why?
Supercooling is observed when a liquid remains liquid even below its normal freezing point. In this case, the lowest temperatures observed for both sucrose solution (-7.9°C) and water (-8.0°C) are well below their freezing points.
- For sucrose, the average freezing point is -1.6°C.
- For water, the average freezing point is 0.0°C.
The difference between the lowest temperature and the average freezing point indicates supercooling:
- For sucrose: [tex]\( -7.9 - (-1.6) = -6.3 \)[/tex]°C
- For water: [tex]\( -8.0 - 0.0 = -8.0 \)[/tex]°C
Thus, we see greater supercooling in the water (-8.0°C) as compared to the sucrose solution (-6.3°C).
### Question 2:
Calculate the number of moles of sugar used.
The molar mass of sugar is given as 342 g/mol, and we used 19 g of sugar.
[tex]\[ \text{Moles of sugar} = \frac{\text{mass of sugar}}{\text{molar mass of sugar}} = \frac{19 \text{ g}}{342 \text{ g/mol}} \approx 0.05556 \text{ mol} \][/tex]
### Question 3:
Calculate the molality of the solution.
Molality (m) is defined as moles of solute per kilogram of solvent. We dissolved 19 g of sugar in 50 mL (0.050 kg) of water.
[tex]\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05556 \text{ mol}}{0.050 \text{ kg}} = 1.111 \text{ mol/kg} \][/tex]
### Question 4:
From the molality, what should the freezing point be?
Freezing point depression can be calculated using the formula:
[tex]\[ \Delta T_f = K_f \times m \][/tex]
Where:
- [tex]\(\Delta T_f\)[/tex] is the freezing point depression.
- [tex]\(K_f\)[/tex] is the cryoscopic constant or freezing point depression constant for water, which is approximately [tex]\(1.86 \text{ °C/mol/kg}\)[/tex].
Therefore:
[tex]\[ \Delta T_f = 1.86 \text{ °C/mol/kg} \times 1.111 \text{ mol/kg} = 2.066 \text{ °C} \][/tex]
Since the normal freezing point of water is [tex]\(0.0 \text{ °C}\)[/tex], the expected freezing point of the sucrose solution would be:
[tex]\[ 0.0 \text{ °C} - 2.066 \text{ °C} = -2.066 \text{ °C} \][/tex]
To summarize:
1. Supercooling was observed, with the water showing greater supercooling than the sucrose solution.
2. The moles of sugar used were approximately [tex]\(0.05556\)[/tex] mol.
3. The molality of the solution was [tex]\(1.111 \text{ mol/kg}\)[/tex].
4. The expected freezing point of the sucrose solution is [tex]\(-2.066 \text{ °C}\)[/tex].
Table 6.1
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text {Solution} & \text {Lowest Temperature (}^{\circ} \mathrm{C}) & \text {1st Trial (}^{\circ} \mathrm{C}) & \text {2nd Trial (}^{\circ} \mathrm{C}) & \text {Average (}^{\circ} \mathrm{C}) & \Delta T_1 (^{\circ} \mathrm{C}) \\ \hline \text {Sucrose} & -7.9 & -1.6 & -1.6 & -1.6 & -7.9 - (-1.6) = -6.3 \\ \hline \mathrm{H}_2 \mathrm{O} & -8.0 & 0.0 & 0.0 & 0.0 & -8.0 - 0.0 = -8.0 \\ \hline \end{array} \][/tex]
### Question 1:
Was there supercooling? Would you expect the water or the sugar solution to have the most supercooling? Why?
Supercooling is observed when a liquid remains liquid even below its normal freezing point. In this case, the lowest temperatures observed for both sucrose solution (-7.9°C) and water (-8.0°C) are well below their freezing points.
- For sucrose, the average freezing point is -1.6°C.
- For water, the average freezing point is 0.0°C.
The difference between the lowest temperature and the average freezing point indicates supercooling:
- For sucrose: [tex]\( -7.9 - (-1.6) = -6.3 \)[/tex]°C
- For water: [tex]\( -8.0 - 0.0 = -8.0 \)[/tex]°C
Thus, we see greater supercooling in the water (-8.0°C) as compared to the sucrose solution (-6.3°C).
### Question 2:
Calculate the number of moles of sugar used.
The molar mass of sugar is given as 342 g/mol, and we used 19 g of sugar.
[tex]\[ \text{Moles of sugar} = \frac{\text{mass of sugar}}{\text{molar mass of sugar}} = \frac{19 \text{ g}}{342 \text{ g/mol}} \approx 0.05556 \text{ mol} \][/tex]
### Question 3:
Calculate the molality of the solution.
Molality (m) is defined as moles of solute per kilogram of solvent. We dissolved 19 g of sugar in 50 mL (0.050 kg) of water.
[tex]\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05556 \text{ mol}}{0.050 \text{ kg}} = 1.111 \text{ mol/kg} \][/tex]
### Question 4:
From the molality, what should the freezing point be?
Freezing point depression can be calculated using the formula:
[tex]\[ \Delta T_f = K_f \times m \][/tex]
Where:
- [tex]\(\Delta T_f\)[/tex] is the freezing point depression.
- [tex]\(K_f\)[/tex] is the cryoscopic constant or freezing point depression constant for water, which is approximately [tex]\(1.86 \text{ °C/mol/kg}\)[/tex].
Therefore:
[tex]\[ \Delta T_f = 1.86 \text{ °C/mol/kg} \times 1.111 \text{ mol/kg} = 2.066 \text{ °C} \][/tex]
Since the normal freezing point of water is [tex]\(0.0 \text{ °C}\)[/tex], the expected freezing point of the sucrose solution would be:
[tex]\[ 0.0 \text{ °C} - 2.066 \text{ °C} = -2.066 \text{ °C} \][/tex]
To summarize:
1. Supercooling was observed, with the water showing greater supercooling than the sucrose solution.
2. The moles of sugar used were approximately [tex]\(0.05556\)[/tex] mol.
3. The molality of the solution was [tex]\(1.111 \text{ mol/kg}\)[/tex].
4. The expected freezing point of the sucrose solution is [tex]\(-2.066 \text{ °C}\)[/tex].
