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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.

Answer :

Explanation:

Formula to calculate refractive index of n is as follows.

     n = [tex]X_{A} \times n_{A} + X_{B} \times n_{B}[/tex]

where,    [tex]X_{A}[/tex] = mole fraction of A

              [tex]X_{B}[/tex] = mole fraction of B

              [tex]n_{A}[/tex] = refractive index of A

              [tex]n_{B}[/tex] = refractive index of B

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

         n = [tex]X_{A} \times n_{A} + X_{B} \times n_{B}[/tex]

         1.5248 = [tex]X_{A} \times 1.7058 + X_{B} \times 1.3658[/tex] ........ (1)

Also, it is known that [tex]X_{A} + X_{B}[/tex] = 1

and,          [tex]X_{B} = 1 - X_{A}[/tex] ......... (2)

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

        1.5248 = [tex]X_{A} \times 1.7058 + 1 - X_{A} \times 1.3658[/tex]  

        1.5248 = [tex]X_{A} \times 1.7058 + 1.3658 - 1.3658X_{A}[/tex]  

         [tex]X_{A}[/tex] = 0.467

And, the value of  [tex]X_{B}[/tex] is calculated as follows.

            [tex]X_{B} = 1 - X_{A}[/tex]

                       = [tex]1 - 0.467[/tex]

                       = 0.533

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

      Mole % of A = [tex]\frac{n_{A}}{n_{A} + n_{B}}[/tex]      

                            = [tex]\frac{0.467}{0.467 + 0.533} \times 100[/tex]

                            = 46.7%

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

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