Assuming 100% dissociation, calculate the freezing point and boiling point of 1.96 m SnCl4 (aq)
delta T = i*Kf*m
i = 5 from 5 particles of SnCl4.
m = molality = 1.96
Kf = constant for water = 1.86 degrees C/m
delta T = 5*1.86*1.96
new freezing point for water is zero C - delta T.
delta T = i*Kb*m
Kb = 0.512 degrees C/m for water
Substitute into the equation and solve for delta T.
new boiling point is 100 + delta T
To calculate the freezing point and boiling point of a solution, you need to use the concept of colligative properties. Colligative properties are properties that depend on the concentration of solute particles in a solution, regardless of their chemical nature. Freezing point depression and boiling point elevation are two such colligative properties.
First, we need to determine the number of particles that dissociate from one formula unit of SnCl4 when it is dissolved in water. SnCl4 is an ionic compound, so it dissociates into Sn4+ ions and 4 Cl- ions.
Next, we need to determine the molality of the solution. Molality (m) is defined as the number of moles of solute divided by the mass of the solvent in kilograms. In this case, the solute is SnCl4, and the solvent is water.
The molar mass of SnCl4 is:
Sn: 118.71 g/mol
C: 35.45 g/mol
So, the molar mass of SnCl4 is:
118.71 g/mol + 4 * 35.45 g/mol = 238.61 g/mol
Since we have a 1.96 m solution, it means that we have 1.96 moles of SnCl4 dissolved in 1 kg of water.
Now, we can calculate the total number of particles in the solution. Since SnCl4 dissociates into one Sn4+ ion and four Cl- ions, the total number of particles will be:
1 * (1.96 mol) + 4 * (1.96 mol) = 9.8 mol
Now, we can use the following equations to calculate the freezing point depression (ΔTf) and boiling point elevation (ΔTb) of the solution:
ΔTf = Kf * m
ΔTb = Kb * m
Where Kf and Kb are the cryoscopic and ebullioscopic constants, respectively, and m is the molality of the solution.
For water, the cryoscopic constant (Kf) is 1.86 °C/m, and the ebullioscopic constant (Kb) is 0.52 °C/m.
Plugging in the values, we have:
ΔTf = (1.86 °C/m) * (1.96 m) = 3.65 °C
ΔTb = (0.52 °C/m) * (1.96 m) = 1.02 °C
Finally, we can use these values to calculate the freezing point and boiling point of the solution:
Freezing point = Freezing point of pure solvent - ΔTf
Boiling point = Boiling point of pure solvent + ΔTb
The freezing point of pure water is 0 °C, so the freezing point of the solution will be:
0 °C - 3.65 °C = -3.65 °C
The boiling point of pure water is 100 °C, so the boiling point of the solution will be:
100 °C + 1.02 °C = 101.02 °C
Therefore, assuming 100% dissociation, the freezing point of 1.96 m SnCl4 (aq) is -3.65 °C, and the boiling point is 101.02 °C.