what is the freezing point of 16.7g of LiCl in 136g of water, assuming complete dissociation for water kf= -1.86/mol
To find the freezing point of 16.7g of LiCl in 136g of water, we need to use the equation for the freezing point depression:
ΔTf = -Kf * m * i
Where:
ΔTf is the freezing point depression (difference between the freezing point of the solvent and the freezing point of the solution),
Kf is the cryoscopic constant (freezing point depression constant),
m is the molality of the solute (amount of solute in moles per 1kg of solvent), and
i is the van't Hoff factor (the number of particles into which the solute dissociates in solution).
First, let's calculate the molality (m) of the LiCl in the solution:
Molar mass of LiCl = 6.94g/mol (Li) + 35.45g/mol (Cl) = 42.39g/mol
Moles of LiCl = (16.7g) / (42.39g/mol) ≈ 0.3948 mol
Mass of water = 136g
Molality (m) = (moles of solute) / (mass of water in kg)
= (0.3948 mol) / (0.136 kg)
≈ 2.903 mol/kg
Now, we need to find the van't Hoff factor (i). Since LiCl is an ionic compound, it completely dissociates into ions in water. Therefore, the van't Hoff factor for LiCl is 2, as it produces two ions (Li+ and Cl-) when it dissolves.
Plug the values into the equation:
ΔTf = -Kf * m * i
= -1.86 °C/mol * 2.903 mol/kg * 2
= -10.803 °C
Therefore, the freezing point of the solution is 10.803 °C lower than the freezing point of pure water.