An aqueous solution of calcium chloride boils at 101.3. How many kilograms of calcium chloride were dissolved in 1000 g of the solvent?
0.0936
Well, if an aqueous solution of calcium chloride boils at 101.3 degrees, it must have really high aspirations. Maybe it's trying to become a professional grade chef and make a mean soufflé! But jokes aside, let's focus on the question.
To find out how many kilograms of calcium chloride were dissolved in 1000 grams of the solvent, we need a little bit of chemistry knowledge.
First, we need to know the boiling point elevation constant, Kb, for the solvent used in this solution. Once we have that value, we can use the formula:
∆Tb = Kb × m
Where ∆Tb is the boiling point elevation, Kb is the boiling point elevation constant, and m is the molality of the solution.
Given that calcium chloride is dissolved in the solvent, we can assume that the boiling point elevation (∆Tb) is 101.3 degrees.
So, unfortunately, without the value for Kb, we can't determine the molality (m) of the solution, and hence we can't calculate how many kilograms of calcium chloride were dissolved.
But hey, at least we had a good laugh with those ambitious calcium chloride molecules, right?
To calculate the amount of calcium chloride dissolved in the solvent, we need to determine the concentration of the solution. The boiling point elevation equation can be used:
ΔTb = Kb * m
Where:
ΔTb = boiling point elevation (difference between the boiling point of the solution and the boiling point of the pure solvent)
Kb = molal boiling point elevation constant
m = molality of the solution (moles of solute per kilogram of solvent)
In this case, we are given ΔTb = 101.3 (boiling point elevation). The molal boiling point elevation constant (Kb) for water is typically 0.512 °C/m.
Let's rearrange the equation to solve for molality (m):
m = ΔTb / Kb
m = 101.3 / 0.512 = 197.851 mol/kg
Now we can calculate the number of moles of calcium chloride dissolved in 1000 g of the solvent by multiplying the molality by the mass of the solvent:
moles of solute = m * mass of solvent
mass of solvent = 1000 g
moles of solute = 197.851 mol/kg * 1 kg = 197.851 mol
Calcium chloride has a molar mass of 111 g/mol.
Finally, we can calculate the mass of calcium chloride dissolved in the solvent by multiplying the moles of solute by the molar mass:
mass of calcium chloride = moles of solute * molar mass
mass of calcium chloride = 197.851 mol * 111 g/mol = 21,941.961 g
Therefore, approximately 21,941.961 grams of calcium chloride were dissolved in 1000 g of the solvent.
To find the number of kilograms of calcium chloride dissolved in 1000 g of the solvent, we need to use the boiling point elevation formula.
The boiling point elevation is given as 101.3°C, which is the boiling point of the solution. The boiling point elevation is determined by the moles of the solute (calcium chloride) dissolved in the solution.
The boiling point elevation (ΔTb) can be calculated using the formula: ΔTb = Kb * m
Here, Kb is the molal boiling point elevation constant (a property of the solvent) and m is the molality of the solution.
Since the boiling point elevation is provided in degrees Celsius, we need to convert it to Kelvin by adding 273.15.
101.3 + 273.15 = 374.45 K
Now, let's calculate the molality of the solution. The molality (m) is defined as the number of moles of solute per kilogram of solvent.
First, we have to convert the given mass of the solvent from grams to kilograms:
1000 g = 1000/1000 = 1 kg
Next, we need to find the number of moles of calcium chloride dissolved in the solution. We can use the formula:
n = m/M
Here, n represents the number of moles, m is the mass of calcium chloride in grams, and M is the molar mass of calcium chloride. The molar mass of calcium chloride is the sum of the atomic masses of calcium (40.08 g/mol) and chlorine (35.45 g/mol):
M = 40.08 + 35.45 = 75.53 g/mol
Substituting the values into the formula, we get:
n = m/M
n = ?/75.53
To find the value of m, we need to use the formula for boiling point elevation:
ΔTb = Kb * m
Since the molality (m) is unknown, we rearrange the formula as:
m = ΔTb / Kb
We need the molal boiling point elevation constant (Kb) for water, which is 0.512 °C/m.
Let's plug in the values:
m = 374.45 / 0.512
Finally, we compute the number of moles by further substituting and rearranging:
n = m/M
?/75.53 = 374.45 / 0.512
To find the value of ?, we can cross-multiply and solve for ?.
CaCl2 is three particles per molecule.
deltaTemp=1.3=.52*molality
solve for molality, then
molality= moles/1000 solve for moles, then finally,grams of CaCl2 from moles, then divide by 1000 to get kg