How many grams of KCl should be added to 1.5kg of water to lower its freezing point to -7.5°C
To calculate the amount of KCl needed to lower the freezing point of water, we can use the equation:
∆Tf = Kf × m
Where:
∆Tf = change in freezing point (in °C)
Kf = cryoscopic constant for water (1.86 °C kg/mol)
m = molality of the KCl solution (moles of solute per kilogram of solvent)
First, let's calculate the change in freezing point (∆Tf):
∆Tf = -7.5°C - 0°C
∆Tf = -7.5°C
Next, we can calculate the molality (m) of the KCl solution. Molality is defined as the moles of solute (KCl) per kilogram of solvent (water).
m = moles of KCl / kilograms of water
Given that we have 1.5kg of water, we can convert it to grams:
1.5kg = 1500g
To calculate the moles of KCl, we need to know the molar mass of KCl:
Molar mass of KCl = atomic mass of K (39.1 g/mol) + atomic mass of Cl (35.45 g/mol)
Molar mass of KCl = 74.55 g/mol
Now, we can calculate the number of moles of KCl:
moles of KCl = (mass of KCl / molar mass of KCl)
Since we want to find the amount of KCl required to lower the freezing point to -7.5°C, we can rearrange the equation ∆Tf = Kf × m to solve for m:
m = ∆Tf / Kf
Now, let's substitute the values into the equation:
m = -7.5°C / 1.86 °C kg/mol
m = -4.03 mol/kg
Finally, we can calculate the mass of KCl needed using the molality (m) and the mass of water:
mass of KCl = m × mass of water
mass of KCl = -4.03 mol/kg × 1500g
mass of KCl = -6045 g
However, the negative sign doesn't make sense in this context since we need a positive mass of KCl. Therefore, we'll take the absolute value of the result:
mass of KCl = 6045 g
So, approximately 6045 grams of KCl should be added to 1.5 kg of water to lower its freezing point to -7.5°C.
To solve this problem, we need to use the equation for calculating the freezing point depression using the molal freezing point constant (Kf) and the molality of the solute:
ΔT = Kf * m * i
where:
ΔT is the change in freezing point
Kf is the molal freezing point constant
m is the molality of the solute
i is the van't Hoff factor (number of particles the solute dissociates into)
First, we need to calculate the molality (m) of the solute. Molality is defined as the moles of solute per kilograms of solvent. In this case, the solvent is water.
First, convert the mass of water from kilograms to grams:
1.5 kg of water = 1500 grams of water
Next, we need to calculate the moles of solute (KCl) required to lower the freezing point. To do this, we can use the molar mass of KCl, which is 74.55 g/mol.
Since we want to lower the freezing point, the change in freezing point (ΔT) is given as -7.5°C.
Now, we need the molal freezing point constant (Kf) for water. The Kf for water is 1.86 °C/m.
Lastly, we need the van't Hoff factor (i) for KCl. KCl dissociates into two particles (one K+ ion and one Cl- ion) when it dissolves.
Now we have all the values needed to solve the equation:
-7.5 = 1.86 * m * 2
To find the molality (m), we rearrange the equation:
m = -7.5 / (1.86 * 2)
m = -7.5 / 3.72
m = -2.02 °C/m
Now, using the definition of molality, we can calculate the moles of KCl required:
molality = moles of solute / kilograms of solvent
-2.02 °C/m = moles of KCl / 1.5 kg
moles of KCl = -2.02 °C/m * 1.5 kg
moles of KCl = -3.03 mol
However, since we cannot have a negative number of moles, we can assume the absolute value of the moles:
moles of KCl = 3.03 mol
Finally, we convert the moles of KCl to grams using the molar mass of KCl:
grams of KCl = 3.03 mol * 74.55 g/mol
grams of KCl = 226.36 g
Therefore, approximately 226.36 grams of KCl should be added to 1.5 kg of water to lower its freezing point to -7.5°C.
delta T = 7.5
i, the van't Hoff factor for KCl is 2.
delta T = i*Kf*molality
Substitute for delta T, i, and Kf (1.86) and solve for m.
Then m = mols/kg solvent. Substitute for m and kg solvent; solve for mols.
Then mol = grams KCl/molar mass KCl. You know molar mass and mols, solve for grams.