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London dispersion forces (LDF) are present in all molecules, whether polar or non-polar. Molecules also exhibiting dipole-dipole interactions (in addition to the LDF) must have stronger forces of attraction than those molecules which exhibit only LDF.

Then, why does tetrachloromethane (carbon tetrachloride), which is a non-polar molecule exhibiting only London dispersion forces, have a higher boiling point ($\pu{77 ^\circ C}$) than trichloromethane (chloroform) ($\pu{61 ^\circ C}$) which is a polar molecule, exhibiting dipole-dipole interactions?

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You also need to account for the difference in dispersion forces between the two molecules. Chlorine is much larger than hydrogen. Therefore tetrachloromethane has a larger molecular surface area which increases the intermolecular interaction strength. In this particular case, it outweighs the weak dipole interactions present in trichloromethane.

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tetrachloromethane has a higher mollecular mass than that of trichloromethane i.e tetrachloromethane has a Rmm=154 and trichloromethane has a Rmm=118.5 thus the heavier the mollecule the greater the forces of attraction between the mollecules hence a high boiling point and vice versa

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    $\begingroup$ What about Polarity of molecules? Why are you not considering them as well ?? $\endgroup$ – Soumik Das May 16 '18 at 12:37

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