Question

8. 12 g of sucrose (CzH201, molecular weight 342 g/mol) is dissolved in 1 of form (CHCla, molecular weight 119 g/mol). What is the boiling point for this chloro solution?

given that Kb=3.63

Answer 1

Mass of sucrose = 12 g

Molar mass of sucrose = 342 g

Moles of sucrose = mass of sucrose/molar mass of sucrose

                            = 12/342

                            = 0.0351 mol

Volume of chloroform = 1 L

Density of chloroform = 1.49 kg/L

Mass of chloroform = 1.49 x 1 = 1.49 kg

Molality of sucrose in the solution (m) = mol of sucrose/kg of solvent

                                                            = 0.0351/1.49

                                                            = 0.0236 m

Boiling point elevation = K_b x m

K_b = boiling point elevation constant = 3.63 ^oC/m

Hence,

boiling point elevation = 3.63 x 0.0236

                                    = 0.09 ^oC

Boiling point of pure chloroform = 61.2 ^oC

Hence, the boiling point of the sucrose solution = (61.2 + 0.09) = 61.29 ^oC