A student measures the refractive index of an organic liquid. The value is 1.282

Question

A student measures the refractive index of an organic liquid. The value is 1.2821 at 18 degree C. What is the corrected refractive index? Do impurities raise or lower the refractive index of a liquid? Explain. Is it proper procedure to add the boiling chips down through the condenser once a solution has started to reflux? What is expected to happen this procedure is followed? What main advantage and disadvantage can you predict about fractional distillation? Compare and contrast the techniques of microscale and miniscale distillation in terms of efficiency of distillation. What are the advantages of each? What are the disadvantages of each?

Explanation / Answer

1. By using the equation for corrected refractive index:

           n20D = nobs + 0.00045(t-20)

Where n20D is the corrected refractive index (D represents the D line of sodium lamp at 20 oC) and nobs is the observed refractive index at temperature t from the refractometer. Thus,

  n20D = 1.2821 + 0.00045 (18-20)

          = 1.2821 + 0.00045 (-2)

         = 1.2812

2. The refractive index of a substance is a measure of the speed of light in that substance. It is expressed as a ratio of the speed of light in vacuum to that in medium.

        Refractive index, nD = speed of light in a vacuum / speed of light in medium

Typically impurity in a substance increases its density and reduces the melting point of a medium (e.g. by decreasing the kinetic energy of particles). A denser liquid has lower temperature and has a higher viscosity resulting light to travel slower in the medium. Thus the larger value of the refractive index is obtained.

Therefore the refractive index is depends on the given impurity which may increase or decrease the density of medium which may raise or lower or unchanged the refractive index value.

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