Final answer:
The freezing point of a solution with 9.2 g of KNO3 in 300 mL of water is -1.13°C. This is calculated by determining the molality of the solution, considering KNO3's dissociation into two ions, and using the freezing point depression formula with water's molal freezing point depression constant.
Step-by-step explanation:
To find the freezing point of a solution consisting of 9.2 g of potassium nitrate (KNO3) dissolved in 300 mL of water, we need to calculate the molality of the solution and then apply the freezing point depression formula. The molar mass of KNO3 is approximately 101.1 g/mol. First, we calculate the number of moles of solute:
Moles of KNO3 = (9.2 g) / (101.1 g/mol) = 0.091 mol
Since the density of water is 1 g/mL, 300 mL of water is equivalent to 300 g. Molality (m) is moles of solute per kilogram of solvent, so:
Molality (m) = 0.091 mol / (0.300 kg) = 0.303 mol/kg
KNO3 dissociates into K+ and NO3- ions, so the van't Hoff factor (i), which is the number of particles the compound dissociates into, is 2. The formula for freezing point depression (ΔTf) is:
ΔTf = i * Kf * m
Insert the values into the equation:
ΔTf = 2 * 1.86°C kg/mol * 0.303 mol/kg = 1.13°C
The freezing point depression is 1.13°C, so the solution will freeze at:
Freezing point of solution = 0°C - 1.13°C = -1.13°C