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%.

it's all four options.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.