Solution.
"Pm = \\frac{Na}{3E0}(\\alpha + \\frac{\\mu^2}{3kT})"
For CCl4 "\\mu" = 0, so
"Pm = \\frac{Na \\times \\alpha}{3E0}"
Pm does not depend on temperature in the case when the dipole moment is 0. Therefore, the A molecule is CCl4. The B - CHCl3 Molecule.
Answer:
Therefore, the A molecule is CCl4. The B - CHCl3 Molecule.
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