Asample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives have passed since the sample originally formed?

3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of radioisotope follows:

By the stoichiometry of above reaction,

1 mole of is produced by 1 mole

So, 0.17 moles of will be produced by =

Amount of

decomposed will be = 0.17 moles

Initial amount of  will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

where,

a = amount of reactant left after n-half lives = 0.024

= Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

Therefore, 3 half lives have passed.

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