Calcium chloride has been used to melt ice from roadways. Given that the saturated solution is 32% calcium chloride by mass, estimate the freezing point. Express your answer in degrees Celsius. Assume that calcium chloride behaves ideally in solution.
32% CaCl2 means 32 g CaCl2/100 g soln or
32 g CaCl2 in 68 g water. The moles CaCl2 then is 32g/molar mass CaCl2
molality = moles CaCl2/0.068 kg.
delta T = i*Kf*m
i = 3. solve for delta T and convert to freezing point.
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To estimate the freezing point of a solution, we need to use the concept of the freezing point depression.
The freezing point depression is given by the equation:
ΔT = Kf * i * m
where ΔT is the change in freezing point, Kf is the cryoscopic constant, i is the van't Hoff factor, and m is the molality of the solution.
In this case, since we are given the mass percent composition instead of the molality, we need to convert it to molality first.
Given that the saturated solution is 32% calcium chloride by mass, it means that for 100g of the solution, there are 32g of calcium chloride:
mass of calcium chloride = 32 g
To convert this mass to moles, we need to know the molar mass of calcium chloride, which consists of one calcium atom (Ca) with a molar mass of 40.08g/mol and two chlorine atoms (Cl) with a molar mass of 35.45g/mol.
molar mass of calcium chloride = 40.08 g/mol + 2*(35.45 g/mol) = 110.98 g/mol
Using the molar mass of calcium chloride, we can calculate the number of moles:
moles of calcium chloride = (mass of calcium chloride) / (molar mass of calcium chloride)
moles of calcium chloride = 32 g / 110.98 g/mol = 0.2881 mol
Next, we need to calculate the molality, which is defined as moles of solute per kilogram of solvent. Since water is typically the solvent in these types of solutions, we assume 1kg of water:
molality = (moles of calcium chloride) / (mass of water in kg)
molality = (0.2881 mol) / (1 kg) = 0.2881 mol/kg
Now, we can calculate the change in freezing point:
ΔT = Kf * i * m
The cryoscopic constant for water is 1.86 °C/m.
i represents the van't Hoff factor, which represents the number of particles produced when the solute dissolves in the solvent. For calcium chloride, it dissociates into one calcium ion (Ca^2+) and two chloride ions (2Cl^-), making the van't Hoff factor i = 3.
Plugging in the values, we have:
ΔT = (1.86 °C/m) * 3 * (0.2881 mol/kg)
ΔT = 1.6465 °C
Finally, to estimate the freezing point, we need to subtract the change in freezing point from the normal freezing point of pure water, which is 0 °C:
Freezing point = 0 °C - 1.6465 °C
Freezing point = -1.6465 °C
Therefore, the estimated freezing point of the calcium chloride solution is approximately -1.6465 °C.
To estimate the freezing point of a solution, we need to use the concept of freezing point depression. This phenomenon is governed by Raoult's Law, which states that the vapor pressure of a volatile component in a solution is directly proportional to its mole fraction.
To begin, we need to convert the given mass percentage of calcium chloride (CaCl2) into a mole fraction. Since the solution is saturated, it means that 32 grams of calcium chloride is dissolved in 100 grams of solution:
Mass of CaCl2 = 32 grams
Mass of solution = 100 grams
To calculate the mole fraction (X) of calcium chloride, we need to determine the moles of calcium chloride (nCaCl2) and moles of solvent (nSolvent). Since calcium chloride is the only solute, the moles of solvent will correspond to water (H2O):
Molar mass of CaCl2 = 40.08 + 2(35.45) = 110.98 g/mol
Moles of CaCl2 (nCaCl2) = Mass of CaCl2 / Molar mass of CaCl2
= 32 g / 110.98 g/mol
≈ 0.288 mol
Molar mass of H2O = 2(1.00784) + 15.999 g/mol = 18.015 g/mol
Moles of H2O (nH2O) = Mass of H2O / Molar mass of H2O
= (100 g - 32 g) / 18.015 g/mol
≈ 3.995 mol
Now, we can calculate the mole fraction of calcium chloride (X):
X = nCaCl2 / (nCaCl2 + nH2O)
= 0.288 mol / (0.288 mol + 3.995 mol)
≈ 0.06712
The mole fraction of water can be obtained by subtracting X from 1:
Mole fraction of H2O = 1 - X
= 1 - 0.06712
≈ 0.93288
Given that calcium chloride behaves ideally in solution, the vapor pressure lowering is directly proportional to the mole fraction of solute particles. Since calcium chloride dissociates into three ions in water (Ca2+ and 2Cl-), we can consider its van't Hoff factor (i) as 3.
Now, we can use the formula for freezing point depression (ΔTf):
ΔTf = i * Kf * molality
Where:
ΔTf = freezing point depression
i = van't Hoff factor
Kf = cryoscopic constant (freezing point depression constant) for the solvent
molality = concentration of the solute in moles of solute per kilogram of solvent
For water, the Kf value is 1.86 °C/m.
To find the molality (m), we need to determine the kilograms of solvent. The mass of water (H2O) in kilograms can be found by converting grams to kilograms:
Mass of H2O = 100 g - 32 g
= 68 g
= 0.068 kg
molality (m) = Moles of solute / Mass of solvent in kg
= nCaCl2 / Mass of H2O in kg
= 0.288 mol / 0.068 kg
≈ 4.235 mol/kg
Finally, we can substitute the values into the freezing point depression formula:
ΔTf = 3 * 1.86 °C/m * 4.235 mol/kg
≈ 24.881 °C
Therefore, the estimated freezing point of the saturated solution of calcium chloride is approximately -24.881 °C.