Number 2 and 3 only. TASK 1: SOLVE MY PROBLEMS!Let us further deepen our understanding by solving the following problems:1. Calculate the boiling point of a solution containing 12g glucose C,H12O6 dissolved in200g of water H2O. (0.5120C – Kp of H2O)2. Calculate the normal freezing point of a 0.7439 mol aqueous solution of C12H22O11 thathas a density of 1.35 g/ml. (C12H22O11 is a non-volatile non-dissociating solute.) Themolal freezing point depression constant of water is 1.86°C kg/mole.3. When 0.279 g of a molecular compound, benzoic acid, was dissolved in 43.0 g ofbenzene, the freezing point of the solution was lowered to 5.15 °C. What is themolecular weight of the benzoic acid? (Note: K; for benzene = 5.12°C/m; Freezing pointof benzene= 5.5°C)

2) Given, [sucrose] = 0.7439 M Density of solution = 1.35 g/ mL # Let the…

2. Calculate the normal freezing point of a 0.7439 mol aqueous solution of C12H22O11 that has a density of 1.35 g/ml. (C12H2½O11 is a non-volatile non-dissociating solute.) The molal freezing point depression constant of water is 1.86°C kg/mole. 3. When 0.279 g of a molecular compound, benzoic acid, was dissolved in 43.0 g of benzene, the freezing point of the solution was lowered to 5.15 °C. What is the molecular weight of the benzoic acid? (Note: Kf for benzene = 5.12°C/m; Freezing point of benzene= 5.5°C)