Freezing point lauric acid = 44.5 degrees CÂÂ
Freezing point solution = 40.7 degrees C
Mass of unknown solid = 0.4156g
Mass of lauric acid = 3.936gÂÂ
The freezing point depression constant of lauric acid is 4.40C/molal
CalculateÂÂ
a) Freezing point depression
b) Molality of unknown (m)
c) Moles of unknown (mol)
d) Molar Mass of unknownÂÂ
The Correct Answer and Explanation is :
Let’s solve step-by-step.
Given Data:
- Freezing point of pure lauric acid ((T_f^0)) = (44.5^\circ \mathrm{C})
- Freezing point of solution ((T_f)) = (40.7^\circ \mathrm{C})
- Freezing point depression constant ((K_f)) = (4.40 \, \mathrm{^\circ C/molal})
- Mass of unknown solute ((m_{\text{solute}})) = (0.4156 \, \mathrm{g})
- Mass of lauric acid ((m_{\text{solvent}})) = (3.936 \, \mathrm{g})
a) Freezing Point Depression ((\Delta T_f))
[
\Delta T_f = T_f^0 – T_f
]
[
\Delta T_f = 44.5^\circ \mathrm{C} – 40.7^\circ \mathrm{C} = 3.8^\circ \mathrm{C}
]
b) Molality of the solution ((m))
The relationship between freezing point depression and molality is:
[
\Delta T_f = K_f \cdot m
]
Rearranging for (m):
[
m = \frac{\Delta T_f}{K_f}
]
[
m = \frac{3.8 \, \mathrm{^\circ C}}{4.40 \, \mathrm{^\circ C/molal}} = 0.8636 \, \mathrm{mol/kg}
]
c) Moles of solute ((n_{\text{solute}}))
The molality formula is:
[
m = \frac{n_{\text{solute}}}{\text{mass of solvent in kg}}
]
Rearranging for (n_{\text{solute}}):
[
n_{\text{solute}} = m \cdot \text{mass of solvent in kg}
]
[
\text{Mass of solvent in kg} = \frac{3.936}{1000} = 0.003936 \, \mathrm{kg}
]
[
n_{\text{solute}} = 0.8636 \, \mathrm{mol/kg} \cdot 0.003936 \, \mathrm{kg} = 0.003398 \, \mathrm{mol}
]
d) Molar Mass of Unknown ((M_{\text{solute}}))
The molar mass is given by:
[
M_{\text{solute}} = \frac{m_{\text{solute}}}{n_{\text{solute}}}
]
[
M_{\text{solute}} = \frac{0.4156 \, \mathrm{g}}{0.003398 \, \mathrm{mol}} = 122.3 \, \mathrm{g/mol}
]
Summary of Results:
a) (\Delta T_f = 3.8^\circ \mathrm{C})
b) (m = 0.8636 \, \mathrm{mol/kg})
c) (n_{\text{solute}} = 0.003398 \, \mathrm{mol})
d) (M_{\text{solute}} = 122.3 \, \mathrm{g/mol})
Explanation (300 words)
The process involves determining the properties of the solution using the freezing point depression principle. First, the freezing point depression ((\Delta T_f)) is calculated by subtracting the solution’s freezing point from that of the pure solvent. This gives (\Delta T_f = 3.8^\circ \mathrm{C}).
Using the given freezing point depression constant ((K_f = 4.40 \, \mathrm{^\circ C/molal})), the molality ((m)) of the solution is calculated. Molality is directly proportional to (\Delta T_f), so dividing (\Delta T_f) by (K_f) yields (m = 0.8636 \, \mathrm{mol/kg}).
To find the number of moles of solute ((n_{\text{solute}})), the definition of molality is used: (m = n_{\text{solute}} / \text{mass of solvent (kg)}). Rearranging and substituting the solvent’s mass in kilograms gives (n_{\text{solute}} = 0.003398 \, \mathrm{mol}).
Finally, the molar mass of the solute ((M_{\text{solute}})) is determined using the formula (M_{\text{solute}} = m_{\text{solute}} / n_{\text{solute}}). Substituting the solute’s mass and the number of moles provides (M_{\text{solute}} = 122.3 \, \mathrm{g/mol}).
This value indicates the unknown compound’s molecular weight, derived using colligative properties, which depend solely on particle concentration, not their identity.