If 45.0 g of C6H12O6, are dissolved in 215 g of H2O, what will be the boiling point of the resulting solution?

(Kb= 0.52oC/m)

A) 100.00oC
B) 100.22oC
C) 100.60oC
D) 112.35oC

C. 100.60oC

Explanation:

Find the molality of the solution first, 1.16 m. Then, multiply by the Kb of 0.52. Then, add that number to the boiling point of pure water (100o). This is the new boiling point of the solution 100.60o.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole