What is the new freezing point of a solution prepared by dissolving 2.25 grams of benzene, C6H6, in 1 85 grams of chloroform? (Freezing Point Chloroform =-65.5 (Kr of Chloroform = 4.67 °C/m) a. the new boiling point of a solution prepared by dissolving 0.925 grams of yellow sulfur, S8, in 110 grams of acetic acid. HC2H02? (Boiling Point Acetic Acid = 119ãC) & (Kb of Acetic Acid = 3.08 °C/m) A solution is prepared by dissolving 4.5 grams of a non-electrolyte solute in 185 grams of water. This new aqueous solution now freezes at-041 1ãC. What is the molecular weight in g mol, of this compound? (Freezing Point Water = 0.0°C) & (Kr of water = 1.86 °C/m) A solution is prepared by dissolving 65.3 milligrams (65.3 mg) of a non-electrolyte solute in 9.75 grams of benzene. This new solution now freezes at 5.26 °C. What is the molecular weight, in g/mol, of this compound? (Freezing Point Benzene 5 (Kr of Benzene = 5.06 °C/m) c. d.