A 4.00-gram sample of sodium chloride was added to a beaker containing 40.0 mL 0.033 M lead(II)
nitrate solution. Lead(II) chloride was precipitated according to the following reaction.
2NaCl(s) + Pb(NO3)2 (aq)  PbCl2(s) + 2NaNO3(aq)
Calculate the mass of lead(II) chloride that can be recovered. Pb = 207; Na = 23; Cl = 35.5; N = 14;
O = 16.

2b2t

explanation:

2b2t

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