calculate the freezing point of a solution of a 3.46g of a compound, X, in 160g of benzene. when a separate of X was vaporised, its density was found to be 3.27g/L at 116 degree Celsius and 773 torr. The freezing point of pure benzene is 5.45 degree Celsius, and kf is 5.12 degree Celsius kg/mol
Calculate the freezing point of a soulution of 3.46 of a compoun
I assume comound X is not polar.
Use P*molar mass = dRT to find molar mass.
Then dT = Kf*m
m = mols/kg solvent. You know kg solvent and mols is g/molar mass.
That gives dT. Subtract from normal freezing point to find new f.p.
Post your work if you get stuck.
12
Student
Ashe
To calculate the freezing point of a solution, you can use the formula:
ΔTf = kf * m
where ΔTf is the change in freezing point, kf is the molal freezing point constant, and m is the molality of the solution.
First, we need to calculate the molality (m) of the solution. Molality is defined as the moles of solute divided by the mass of the solvent in kg.
Given:
Mass of compound X = 3.46g
Mass of benzene = 160g
To find the moles of X, we need to first determine its molar mass. Since we don't have the formula or the molar mass of compound X, we can't directly calculate the moles. However, since we have the density and temperature of vaporization, we can calculate its molar mass using the ideal gas law.
The ideal gas law is given by:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Given:
Density of X = 3.27g/L
Temperature = 116°C = 389K (converted to Kelvin)
Pressure = 773 torr (converted to atm)
First, convert the density to the molar mass of X by dividing the density by the molar volume at STP (standard temperature and pressure). The molar volume at STP is 22.4 L/mol.
Molar mass of X = Density of X / Molar volume
= 3.27 g/L / 22.4 L/mol
= 0.146 mol/g
Next, calculate the moles of X using the given mass:
Moles of X = Mass of X / Molar mass of X
= 3.46 g / 0.146 mol/g
= 23.7 mol
Now, calculate the molality of the solution:
Molality (m) = Moles of X / Mass of benzene (in kg)
= 23.7 mol / 0.160 kg
= 148.13 mol/kg
Finally, we can substitute the known values into the freezing point equation:
ΔTf = kf * m
= 5.12 °C kg/mol * 148.13 mol/kg
= 758.54 °C
To find the freezing point of the solution, subtract the change in freezing point from the freezing point of pure benzene:
Freezing point of solution = Freezing point of pure benzene - ΔTf
= 5.45 °C - 758.54 °C
= -753.09 °C
Therefore, the freezing point of the solution is -753.09 °C.