Answer:
Answer = − 3.54°C
Explanation:
Finding the freezing point depression associated with 14.8g of aluminum nitrate we have
Al(NO3)3, in that much water.
Freezing-point depression depends on the number of particles of solute present in solution.
When aluminum nitrate dissociates completely in water we have
Al(NO3)3(s) → Al3+(aq) + 3(NO3)1-(aq)
3(aq]
One mole of aluminium nitrate produces 4 moles of ions in solution, 1 mole of aluminium cations and 3 moles of nitrate anions.
freezing-point depression equation = ΔT f = i × Kf × b,
where ΔT f = freezing-point depression;
i = van Hoff factor
Kf = cryoscopic constant of the solvent;
b = molality of the solution.
The cryoscopic constant of water in the question is equal to Kf = 1.86°Cm= 1.86°C kg mol-1
van Hoff factor, i = 4 which is the number of moles Al(NO3)3(s) dissociated into
The molality of the solution is given by moles of solute / mass of solvent
the number of moles of solute = the mass of the solvent expressed in kilograms
The number of moles of solute is given as = mass/(molar mass)
14.8g/ 212.996 g/mol = 0.0695 moles of Al(NO3)3
The solution’s molality is thus
b = 0.0695 moles/(146×10-3Kg) = 0.476 moles/Kg
Substituting into the freezing-point depression equation we have ΔT f
ΔT f = 4 × 1.86°Ckgmol−1 × 0.476 mol kg−1 = 3.54°C
The solution’s freezing point will be
ΔTf = T0f −T f → T f = T° f− ΔT f Here T°f = freezing point of the pure solvent = 0°C.
T f = 0°C − 3.54°C
Answer = − 3.54°C
