The freezing point for pure lauric acid is 43.8 °C. When 8.00 g of lauric acid is mixed with 1.00 g of benzoic acid, the freezing point drops to 39.2 °C.
What is the experimental value for the molality of the mixture if Kfp for lauric acid is –3.9 °C/m?
What is the number of moles of the solute?
What is the experimental molar mass of the solute?
The Correct Answer and Explanation is :
To solve this, we need to calculate the experimental molality, moles of the solute, and the experimental molar mass of the solute.
Step 1: Calculate the change in freezing point (ΔTf)
The freezing point depression formula is:
[
\Delta T_f = K_f \times m
]
Where:
- (\Delta T_f) is the freezing point depression,
- (K_f) is the freezing point depression constant of the solvent (lauric acid),
- (m) is the molality of the solution.
First, calculate the change in freezing point:
[
\Delta T_f = T_{\text{freeze, solvent}} – T_{\text{freeze, solution}}
]
[
\Delta T_f = 43.8^\circ C – 39.2^\circ C = 4.6^\circ C
]
Step 2: Calculate molality (m)
Now that we have the freezing point depression, we can rearrange the freezing point depression formula to solve for molality (m):
[
m = \frac{\Delta T_f}{K_f}
]
[
m = \frac{4.6^\circ C}{3.9^\circ C/m} = 1.18 \, \text{mol/kg}
]
So, the experimental molality is 1.18 mol/kg.
Step 3: Calculate the moles of solute (benzoic acid)
Next, we use the definition of molality:
[
m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}
]
We know that:
- Molality (m = 1.18 \, \text{mol/kg}),
- Mass of the solvent (lauric acid) = 8.00 g = 0.00800 kg.
Rearranging the equation to solve for moles of solute:
[
\text{moles of solute} = m \times \text{mass of solvent (kg)}
]
[
\text{moles of solute} = 1.18 \, \text{mol/kg} \times 0.00800 \, \text{kg} = 0.00944 \, \text{mol}
]
So, the moles of benzoic acid (solute) are 0.00944 mol.
Step 4: Calculate the experimental molar mass of the solute
Finally, we can calculate the experimental molar mass of the solute using:
[
\text{Molar mass} = \frac{\text{mass of solute (g)}}{\text{moles of solute (mol)}}
]
[
\text{Molar mass} = \frac{1.00 \, \text{g}}{0.00944 \, \text{mol}} = 105.9 \, \text{g/mol}
]
Thus, the experimental molar mass of the solute (benzoic acid) is 105.9 g/mol.
Summary of Results:
- Molality = 1.18 mol/kg
- Moles of solute = 0.00944 mol
- Experimental molar mass of the solute = 105.9 g/mol
This approach uses the concept of freezing point depression and molality to find the molar mass of the solute in the mixture.