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 C? The freezing point for pure water is 0.0 C and Kf is equal to 1.86 C/m.
* Use Tf = Kf*i*m
i=2
m=gramsKNO3/(molmassKNO3*.275)
find grams KNO3 from your fourmula.
To solve this problem, we need to use the equation Tf = Kf * i * m, where Tf is the change in freezing point, Kf is the cryoscopic constant, i is the van't Hoff factor (which corresponds to the number of particles the solute dissociates into), and m is the molality (moles of solute per kilogram of solvent).
First, let's calculate the change in freezing point (Tf):
Tf = -14.5 °C - 0.0 °C
Tf = -14.5 °C
Next, we need to find the molality (m) using the given volume of water (275 mL) and the density of water at room temperature (1 g/mL):
Mass of water = volume * density = 275 mL * 1 g/mL = 275 g
Since we need the molality in kilograms, we divide the mass by 1000:
m = mass of solute / (mass of solvent in kg)
m = 1 / (275 g / 1000) = 1 / 0.275 kg ≈ 3.64 mol/kg
Now, we need to determine the van't Hoff factor (i) for KNO3. The KNO3 dissociates completely in water, producing two ions (K+ and NO3-), so i = 2.
With all the required values, we can rearrange the equation Tf = Kf * i * m to solve for the mass of KNO3:
Tf = Kf * i * m
-14.5 °C = (1.86 °C/m) * 2 * 3.64 mol/kg * mass of KNO3
Solving for the mass of KNO3:
mass of KNO3 = -14.5 °C / [(1.86 °C/m) * 2 * 3.64 mol/kg]
mass of KNO3 ≈ 132.5 g
Therefore, approximately 132.5 grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at -14.5 °C.