Answer:
Freezing point of solution = -2.740C
Explanation:
The solution is attached herewith
The normal freezing point of an aqueous solution of C₁₂H₂₂O₁₁ is -1.26°C
The freezing point depression can be calculated using the formula:
ΔTf = i* Kf*m ; where
The molality of the aqueous solution of C₁₂H₂₂O₁₁ is first determined using the formula:
The mass of water is unknown but is determined from the mass of the solute and solution from the formula:
mass of solute = number of moles * molar mass
moles of solute in 1 L solution = 0.7439 moles
molar mass of solute, C₁₂H₂₂O₁₁ = 342.0 g/mol
mass of solute = 0.7439 moles * 342.0 g
mass of solute = 254.41 g
mass of solution = density * volume
volume of solution = 1 L = 1000 mL
mass of solution = 1.35g/ml * 1000 mL
mass of solution = 1350 g
Therefore, mass of solvent (water) = 1350 g - 254.41 g
mass of solvent (water) = 1095.59 g
mass of solvent (water) in kg = 1095.59 g * 1 kg/ 1000 g
mass of solvent (water) in kg = 1.0956Kg
Thus, molality of solution = 0.7439 moles / 1.0956 Kg
molality of solution = 0.679 mol/kg
The freezing point depression, ΔTf, is then determined using the formula: ΔTf = i* Kf*m
Since C₁₂H₂₂O₁₁ is a non-electrolyte; i for C₁₂H₂₂O₁₁ = 1
Kf of water = 1.86 kg/mole
m = 0.679 mol/kg
ΔTf = 1 * 1.86 * 0.679
ΔTf = 1.26 °C
Therefore, freezing point depression, ΔTf = 1.26 °C
normal freezing point of solution = freezing point of water - freezing point depression
normal freezing point of solution = 0.0°C - 1.26°C
normal freezing point of solution = -1.26°C
Therefore, the normal freezing point of an aqueous solution of C₁₂H₂₂O₁₁ is -1.26°C
Learn more about freezing point and molality at: https://brainly.com/question/25783932
