10.) Freezing point depression (ΔTf)Use the equation:ΔTf = Tf (solution) – Tf (solvent)The freezing point of pure water is 0.00 °C 11.) Actual van't Hoff factor (i) for CaCl2Use the equation:ΔTf = –i Kf mKf for water is 1.86 °C/m 10.97441. Mass CaCl, (8)100.02. Volume of water (mL)51.36113. Mass ice (g)-3.57Freezing point of the4.solution (*C)of//Part 2 CalculationsWhat is the predicted value of i for CaCl,?35.For electrolytes, assume i = the number ofions per formula unit.of//Moles of solute (mol)6.The molar mass of CaCl, is 110.98 g/mol.//Mass of solvent in gramsAdd together the masses of the liquid water7. and ice added to the calorimeter. Convertthe volume of liquid water to mass using1.000 g/ml as the density of water (seeprocedure for volume used).//Mass of solvent in kilograms8.Use the metric definition 1 kg = 1000 g

We have given that 1) Mass of CaCl2 = 10.9744 g 2) Volume of water = 100 mL 3) Mass of ice = 51.3611 g 4) Freezing point of the solution= - 3.57°C 5) Predicted value of i for CaCl2 = 3 And we have asked to find 6) Moles of solute =? 7) Mass of solvent in g =? 8) Mass of solvent in kg =? 10) Freezing point depression (ΔTf) =? 11) Actual van't Hoff factor (i) for CaCl2 =?6) Moles of solute (CaCl2) = Mass of CaCl2Molar mass of CaCl2=10.9744g110.98 g/mole= 0.099 mole 7) Mass of solvent = (Mass of ice) + (Mass of water) = (51.3611g) + (100mL×1g/mL) = 151.3611 g 8) Mass of solvent in kg = 151.3611g×1kg1000g=0.151 kg 10) Freezing point depression (ΔTf) : Tf for pure water = 0.00°C ΔTf = Tf (solution) – Tf (solvent) = - 3.57°C - 0.00°C = - 3.57°C 11) Actual van't Hoff factor (i) for CaCl2: Kf for water is 1.86 °C/m ΔTf = –i Kf m rearrange this equation i i = -ΔTf Kfm molality (m) = Moles of solute Mass of solvent in kg = 0.099 mole0.151 kg= 0.65 m now put all values in above equation i = -ΔTf Kfm= -(-3.57°C)(1.86°C/m)(0.65m)=2.95 ≅ 3