Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at -14.5 degrees celcius ? The freezing point for pure water is 0.0 degrees celcius and Kf is equal to 1.86 degrees celcius/m.
If the 3.90 m solution from Part A boils at 103.45 degrees celcius, what is the actual value of the van't Hoff factor, i? The boiling point of pure water is 100.00 degrees celcius and Kb is equal to 0.512 degrees celcius/m
Did you work the first part to obtain 3.90m for the KNO3 solution. Then
m = moles/kg solvent.
Solve for moles.
Moles = grams/molar mass
Solve for grams. I think it is about 110 g or so but you need to work it through.
Second part.
(103.45-100) = i*0.512*3.90
Solve for i.
To calculate the grams of KNO3 needed to produce a solution that freezes at -14.5 degrees Celsius, we need to use the formula:
∆Tf = i * Kf * molality
1. Convert the given freezing point to Kelvin:
-14.5 degrees Celsius + 273.15 = 258.65 K
2. Convert the volume of water to liters:
275 mL = 0.275 L
3. Calculate the molality, which is the moles of solute per kilogram of solvent:
molality = moles of solute / kilograms of solvent
Since we're assuming complete dissociation, the moles of KNO3 will be equal to the moles of K+ and NO3-, so we can use the molality formula for ions:
molality = (moles of ions) / kilograms of solvent
4. Use the equation for freezing point depression to solve for the molality:
∆Tf = i * Kf * molality,
where ∆Tf = Tf (freezing point of pure solvent) - Tf (freezing point of solution)
0.0 - (-14.5) = i * 1.86 * molality
14.5 = i * 1.86 * molality
5. Rearrange the equation to solve for molality:
molality = 14.5 / (i * 1.86)
6. Convert the mass of water to kilograms:
mass of water = volume of water (in liters) * density of water
Assuming the density of water is approximately 1 g/mL, the mass of 275 mL of water is 275 grams.
mass of water = 275 g = 0.275 kg
7. Solve for moles of KNO3:
moles of KNO3 = molality * kilograms of solvent
8. Calculate the molar mass of KNO3:
molar mass of KNO3 = atomic mass of K + atomic mass of N + 3 * atomic mass of O
molar mass of KNO3 = 39.10 g/mol + 14.01 g/mol + 3 * 16.00 g/mol
9. Find the grams of KNO3 using the equation:
grams of KNO3 = moles of KNO3 * molar mass of KNO3
Now let's move on to the second part of the question concerning the van't Hoff factor:
To find the actual value of the van't Hoff factor (i), we need to use the boiling point elevation formula:
∆Tb = i * Kb * molality
1. Convert the boiling point to Kelvin:
103.45 degrees Celsius + 273.15 = 376.6 K
2. Use the equation for boiling point elevation:
∆Tb = i * Kb * molality
(376.6 - 100.0) = i * 0.512 * molality
276.6 = i * 0.512 * molality
3. Rearrange the equation to solve for i:
i = 276.6 / (0.512 * molality)
Therefore, to determine the actual value of the van't Hoff factor, we need to know the molality of the solution.
To solve the first part of the question, we need to use the freezing point depression equation:
ΔT = i * Kf * m
Where:
ΔT = change in freezing point (in degrees Celsius)
i = van't Hoff factor
Kf = freezing point depression constant (in degrees Celsius/mole)
m = molality of the solution (in moles of solute per kilogram of solvent)
In this case, we have the change in freezing point (ΔT) given as -14.5 degrees Celsius, and the freezing point depression constant (Kf) given as 1.86 degrees Celsius/m. We want to find the mass (in grams) of KNO3 to be added.
First, we need to convert the volume of water to kilograms:
275 mL of water = 0.275 kg of water
Next, we need to calculate the molality (m) using the formula:
m = (moles of solute) / (mass of solvent in kg)
Since we have no information about the number of moles of KNO3, we can't directly calculate the molality. However, we can assume complete dissociation of the solute, so 1 mole of KNO3 will produce 2 moles of ions (K+ and NO3-). Therefore, the moles of solute can be calculated as follows:
moles of solute = i * moles of ions
moles of ions = 2 * moles of KNO3
Given that we don't know the number of moles of KNO3, we will assume a certain value and calculate the moles of ions accordingly. Let's assume we have x moles of KNO3.
moles of ions = 2 * x = 2x
moles of solute = i * 2x
Now we can calculate the molality:
m = (i * 2x) / 0.275 kg
Substituting the given values, we have:
-14.5 = i * 1.86 * ((i * 2x) / 0.275)
Simplifying the equation, we get:
-14.5 = 3.72 * i^2 * x
Now we have an equation with two unknowns (i and x). To solve for one of them, we need additional information. Unfortunately, the problem statement doesn't provide any. Hence, we can't determine the exact grams of KNO3 required without knowing one of the variables.
Moving on to the second part of the question, to find the actual value of the van't Hoff factor (i), we need to use the boiling point elevation equation:
ΔT = i * Kb * m
Where:
ΔT = change in boiling point (in degrees Celsius)
Kb = boiling point elevation constant (in degrees Celsius/mole)
m = molality of the solution (in moles of solute per kilogram of solvent)
In this case, the change in boiling point (ΔT) is given as 103.45 degrees Celsius, and the boiling point elevation constant (Kb) is given as 0.512 degrees Celsius/m. We want to find the actual value of the van't Hoff factor (i).
Using the same logic as before, we can assume complete dissociation of the solute and find the molality (m) as follows:
m = (moles of solute) / (mass of solvent in kg)
Let's assume we have y moles of the solute. Since we have no information about the number of moles of the solute, we will need to assume a certain value:
moles of solute = y
moles of ions = 2 * y
moles of solute = i * 2y
Now, we can calculate the molality:
m = (i * 2y) / 0.275 kg
Substituting the given values, we have:
103.45 = i * 0.512 * ((i * 2y) / 0.275)
Simplifying the equation, we get:
103.45 = 1.86 * i^2 * y
Similarly to the first part of the question, we have an equation with two unknowns (i and y). We can't solve for the actual value of the van't Hoff factor (i) without additional information about the number of moles of solute (y).