The magnitude of the dipole moment of a water molecule in liquidwater is 6.2×10

Question

The magnitude of the dipole moment of a water molecule in liquidwater is 6.2×1030 C·m. Asimplified model of the water molecule H2O can be constructed by imagining that each ofthe two hydrogen atoms in the molecule contributes its singleelectron to the one oxygen atom, and thereby becomes bound to it.In this model, the molecule is a simple dipole, with the oxygenatom at one end and the two hydrogen atoms at the other, each endhaving a charge of magnitude2e = 3.2×1019 C. Inthis model, what is the dipole separation of the charges in thewater molecule?

Explanation / Answer

Hi, From the given explanation it could be understood thatH2O molecule can be considered as a dipole withcharge                                        q= 2 e = 3.2 * 10-19 C .    Also given that dipole moment of thismolecule = 6.2 * 1030 C·m    We know that the dipole moment of a dipoleis given by p = q d   (where q is the charge ateach end                                                                                                             and d is the separation of the two charges).          thus, d= p / q = ( 6.2 * 1030 ) /  ( 3.2* 10-19 ) = 1.9375 * 10-11 m Hope this helps you. Hope this helps you.

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