Question

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 1

Explanation:

Formula to calculate refractive index of n is as follows.

     n = $X_{A} \times n_{A} + X_{B} \times n_{B}$

where,    $X_{A}$ = mole fraction of A

              $X_{B}$ = mole fraction of B

              $n_{A}$ = refractive index of A

              $n_{B}$ = refractive index of B

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

         n = $X_{A} \times n_{A} + X_{B} \times n_{B}$

         1.5248 = $X_{A} \times 1.7058 + X_{B} \times 1.3658$ ........ (1)

Also, it is known that $X_{A} + X_{B}$ = 1

and,          $X_{B} = 1 - X_{A}$ ......... (2)

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

        1.5248 = $X_{A} \times 1.7058 + 1 - X_{A} \times 1.3658$  

        1.5248 = $X_{A} \times 1.7058 + 1.3658 - 1.3658X_{A}$  

         $X_{A}$ = 0.467

And, the value of  $X_{B}$ is calculated as follows.

            $X_{B} = 1 - X_{A}$

                       = $1 - 0.467$

                       = 0.533

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

      Mole % of A = $\frac{n_{A}}{n_{A} + n_{B}}$      

                            = $\frac{0.467}{0.467 + 0.533} \times 100$

                            = 46.7%

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