Question 13.4.7: What is the freezing point of a solution containing 40.0 g I......
What is the freezing point of a solution containing 40.0 g I_{2}, a nonelectrolyte, and 250 g benzene (C_{6}H_{6})?
You are asked to calculate the freezing point of a solution.
You are given the quantity of solute and solvent in the solution.
The freezing point is the normal freezing point minus the freezing point depression. Using data from Table 13.4.2,
T_{fp} = 5.5 °C – (5.12 °C/m)(m_{solute})i
The molality of I_{2} is
m_{I_{2}}=\frac{\text{ mol I}_{2}}{\text{ kg benzene}}=\frac{\left(40.0\text{ g}\right)\left(\frac{1\text{ mol I}_{2}}{253.8 \text{ g}} \right) }{0.250 \text{ kg benzene}} = 0.630 m
Use equation 13.10 to calculate the freezing point of the solution.
ΔT_{fp} = K_{fp}m_{solute}i (13.10)
T_{fp} = 5.5 °C – (5.12 °C/m)(0.630 m)(1) = 2.3 °C
| Table 13.4.2 Freezing Points and Depression Constants for Common Solvents |
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| Solvent | T_{fp} (°C) | K_{fp} (°C/m) |
| Water | 0 | 1.86 |
| Benzene | 5.5 | 5.12 |
| Acetic acid | 16.6 | 3.90 |
| Nitrobenzene | 5.7 | 7.00 |
| Phenol | 43 | 7.40 |
| Camphor | 179.4 | 40.0 |