A2.60 gram sample of a compound know to contain only indium and chlorine is dissolved in 50.0 g of tin(iv) chloride (kb = 9.43oc kg mol-1). the normal boiling point is raised from 114.1oc for pure sncl4 to 116.3oc for the solution. what is the molecular weight and probable molecular formula of the solute?

1. The molecular weight of the compound is 222.8 g/mol

2. The probable molecular formula of the solute is

Explanation:

Elevation in boiling point is given by:

= elevation in boiling point

i= vant hoff factor = 1 (for non electrolyte)

= boiling point constant =

m= molality

Weight of solvent (tin chloride)= 50.0 g =0.05 kg

Molar mass of unknown solute = M g/mol

Mass of unknown solute = 2.6 g

The possible formula for the compound would be as indium has valency of 3 and chlorine has valency of 1 has molecular mass almost equal to 222.8.

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