Question

A solution of glucose (C₆H₁₂O₆) is prepared by dissolving 100.0 g of glucose in 1000. g of water. The density of the resultant solution is 1.050 g/mL. Kb for water is 0.52 ^oC/m and kf for water is –1.86 ^oC/m.

What is the vapor pressure of the solution at 100.0^oC?

What is the boiling point of the solution?

please i really need a clear explanation for this

What is the osmotic pressure of the solution at 25^oC?

Answer 1

(Tb - T0b) = kb * molality of solution

(Tb - 100) = 0.52 * (100/180*1000/1000)

Boiling point of the solution Tb = 100.29

P/P0 = mole fraction of Solvent

P = vapour pressure of solution and P0 = vapour pressure of pure water

P/1 atm = (1000/(100+1000))

P = 0.9091 atm

Osmotic pressure of the solution = C * S * T

C = concentration of solution ( C = mass/molar mass * 1000/ volume of the solution)

S = solution constant, T = absolute teperature
                               
                                 = (100/180*1000/1050) * 0.083 * 373

                                 = 16.38 atm