The formula of the gas ozone is O 3. What is the volume of 48g of ozone at r. t.p?

1.8 x 10^ 24 atoms of oxygen

Explanation:

The molecular weight of ozone is known to be 48 grams / mol. Here we are given a sample of 48 grams of ozone as well, so in 48 grams of ozone the number of moles = 48 / 48 = 1,

1 mole of ozone is equal to 6.0221415 × 10^23 molecules of ozone. Respectively, 1 molecule of ozone has 3 atoms of oxygen. Thus, you can conclude the following -

3 * 6.0221415 × 10^23 =  ( About ) 1.8 x 10^ 24 atoms of oxygen

Hope that helps!

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