Calculate the mass of silver chloride in grams required to make 5.8 g of silver plating.

Silver chloride is simply AgCl while silver is Ag so we see that there is one mole to one mole correspondence. First calculate the moles of Ag:

moles Ag = 5.8 g / ( 107.87 g/mol)

moles Ag = 0.0538 mol

Then the mass of AgCl required:

mass AgCl = 0.0538 mol * 143.32 g/mol

mass AgCl = 7.71 grams

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