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# What is the freezing point of a solution that contains 36.0 g of glucose in 500.0 g of water (Kf for water is 1.86C/m. The molar mass of glucose is 180.0 g per mole.) A 0.755 B -0.744 C 1.49 D-1.49 What is the molality of a solution of water and KCl if the boiling point of the solution is 103.07 degrees celcius. (Kb for water =0.512 C/m, KCl is an ionic compound) A 0.300 B 0.600 C 3.00 D 6.00 PLEASEEE help! I have been having so much trouble with this. Have no idea where to even start!

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1) The freezing point depression is a colligative property.For ionic compounds could not be calculated with the formula:ΔTf = Kf * mWhere Kf is the cryoscopic constant of water = 1.86°C/molAnd m is the molality, m = number of moles of solute / kg of solventYou have 500.0 g of solvent = 0.5000 kgAnd you can calculate the moles in 36.0 g of glucose by dividing by its molar mass:number of moles = mass in grams / molar mass = 36.0 g / 180.0 g/mol = 0.2 mol=> m = 0.2 mol / 0.5000 kg = 0.4 m=> DT = 1.86°C/m * 0.4 m = 0.744 °C. Since the freezing point of pure water is 0.0°C, the new freezing point is 0°C - 0.744 °C = - 0.744°CAnswer: - 0.744 °C2) the Elevation of the boiling point is another colligative property.It is calculated as:ΔTb = i * Kb * m.Where i is van't Hoff constant, which represents the number of generated ions in ionic compounds.Here, you don't have the i constant, but you can tell that each molecule of KCl dissociates into 2 ions (one K+ and one Cl-) i.e. i = 2.Kb is the boiling point constant of the solvent, which here is the water => Kb = 0.512 °C/m.You also know that the boiling point of the solution as 103.07°C. Since the normal boiling point of pure water is 100.0°C, ΔT is 103.07°C - 100.0 °C = 3.07°C. Then, you can determine the molality, m = ΔTb / (i*Kb) = 3.07°C / (2*0.512°C/m) = 3.00 °C. Answer: 3.0 °C