Ask Question
3 June, 19:50

You plan to separate a polar and a non-polar compound using normal phase column chromatography technique. Methylene chloride (CH2Cl2) and hexanes (C6H14) are the solvents to be used to elute the column. Which solvent should be used first

+2
Answers (1)
  1. 3 June, 20:28
    0
    Hexane should be used first.

    Explanation:

    Chromatography is a method of separating the constituents of a mixture by taking advantage of their different rates of movement in a solvent over an adsorbent medium. Chromatography is a means of separation and analysis that utilises fractional separation. It is based on the principle that if a fluid containing a number of substances is allowed to pass though an adsorbent medium, the different substances in the fluid may travel at different rates and be separated.

    The rate of movement depends on the relative affinities of the constituents for the solvent and adsorbent medium. i. e solutes which are weakly adsorbed by the adsorbent medium are easily redissolved by the ascending solvent and quickly travel up the adsorbent medium. In addition to that, solutes which are very soluble in the solvent move up at a faster rate than those which are not soluble.

    In column Chromatography;

    A non-polar solvent should be initiated and applied first. This is because, in a column chromatography, a non-polar compound will be removed at first then later polar compound.

    Assuming a polar compound is used first, the polar compounds will be removed alongside with all the non-polar compounds.

    From the two Compounds given;

    We know that:

    Hexane is a non-polar compound and Methylene chloride is a polar compound. As such, Hexane should be used first.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “You plan to separate a polar and a non-polar compound using normal phase column chromatography technique. Methylene chloride (CH2Cl2) and ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers