A student was trying to determine the mole percent of A in a mixture of A and B using refractive index. If their mixture has a refractive index of 1.5248 and pure A and pure B each had refractive indices of 1.7058 and 1.3658, respectively, what was the mole percent of A in their mixture. Type your numerical answer rounded to the 3rd decimal place (i. e. 45.982 or 9.550, etc) without a percent sign.

Explanation:

Formula to calculate refractive index of n is as follows.

     n =

where,     = mole fraction of A

              = mole fraction of B

              = refractive index of A

              = refractive index of B

Hence, putting the given values into the above formula as follows.

         n =

         1.5248 = ........ (1)

Also, it is known that = 1

and,           ......... (2)

Now, put equation (2) in equation (1) as follows.

        1.5248 =  

        1.5248 =  

          = 0.467

And, the value of   is calculated as follows.

                       =

                       = 0.533

Hence, mole percentage of A will be calculated as follows.

      Mole % of A =      

                            =

                            = 46.7%

Thus, we can conclude that the mole percent of A in their mixture is 46.7%.

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