assuming complete dissociation of the solute, how many grams of \rm KNO_3 must be added to 275 \rm mL of water to produce a solution that freezes at -14.5 ^\circ {\rm C}? The freezing point for pure water is 0.0 ^\circ {\rm C} and K_f is equal to 1.86 ^\circ {\rm C}/m.
delta T = i*Kf*m
Solve for m.
m = moles KNO3/kg solvent
Solve for moles KNO3.
moles KNO3 = grams KNO3/molar mass KNO3
Solve for grams.
i = 2
To solve this problem, we'll use the equation for freezing point depression:
ΔT = i * K_f * molality
Where:
ΔT is the change in freezing point
i is the van't Hoff factor (the number of particles the solute dissociates into)
K_f is the cryoscopic constant (given as 1.86 °C/m)
molality is the molality of the solution (moles of solute per kilogram of solvent)
First, let's calculate the change in freezing point (ΔT):
ΔT = (Freezing point of solvent) - (Freezing point of solution)
ΔT = 0.0 °C - (-14.5 °C)
ΔT = 14.5 °C
Since KNO3 dissociates completely into positive potassium (K+) ions and negative nitrate (NO3-) ions, the van't Hoff factor (i) for KNO3 is 2. This is because it forms two particles (K+ and NO3-) when it dissociates in water.
Next, we need to calculate the molality (m) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent.
We know the volume of water is 275 mL, but we need to convert it to kilograms. The density of water is approximately 1 g/mL or 1000 kg/m^3. Therefore, the mass of water is:
Mass of water = Volume of water * Density of water
Mass of water = 275 mL * 1 g/mL
Mass of water = 275 g
Converting the mass of water to kilograms:
Mass of water = 275 g * (1 kg / 1000 g)
Mass of water = 0.275 kg
Now, let's calculate the molality using the given mass of KNO3:
Molality (m) = (moles of KNO3) / (mass of water in kg)
To determine the moles of KNO3, we need to use its molar mass.
Atomic mass of K = 39.10 g/mol
Atomic mass of N = 14.01 g/mol
Atomic mass of O = 16.00 g/mol
Molar mass of KNO3 = (39.10 g/mol) + (14.01 g/mol) + (3 * 16.00 g/mol)
Molar mass of KNO3 = 101.10 g/mol
Now, let's calculate the moles of KNO3:
moles of KNO3 = (mass of KNO3) / (molar mass of KNO3)
Since we don't know the mass of KNO3 yet, let's represent it as 'x' grams.
moles of KNO3 = (x g) / (101.10 g/mol)
Finally, we can substitute all the values into the equation for molality:
m = (moles of KNO3) / (mass of water in kg)
m = [(x g) / (101.10 g/mol)] / 0.275 kg
Now, we have a value for molality. We can substitute all the values into the freezing point depression equation to solve for 'x':
ΔT = i * K_f * m
14.5 °C = 2 * (1.86 °C/m) * {[(x g) / (101.10 g/mol)] / 0.275 kg}
Solving this equation will give us the mass of KNO3 required to produce the desired freezing point depression.