Determine the molality of an aqueous solution that is 15.5 percent urea by mass.

15.5% by mass is equivalent 15.5 g urea in 100 g solution or 155 g urea in 1 kg solution. 

we know that molality = moles solute / kg solvent

moles solute = 155 g / 60 g/mol = 2.58 moles urea

Since there are 155 g urea in 1000g solution, hence the solvent is 845 g or 0.845 kg

So:
molality = 2.58 / 0.845 = 3.06 m

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