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A neutral water molecule `(H_(2)O)` in its vapour state has an electric dipole moment of magnitudes `6.4xx10^(-30)C-m`. How far apart are the molecules centres of positive and negative chargeA. `4m`B. `4 mm`C. `4 mu m`D. `4 p m` |
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Answer» Correct Answer - d There are `10` electrons and `10` protons in a neutral water molecule. So its dipole moment is `p=q (2l)= 10e(2l)` Hence length of the dipole i.e., distance between centres of positive and negative charge is `2l=(p)/(10e)=(6.4xx10^(-20))/(10xx1.6xx10^(-19))= 4xx10^(-12)m= 4p m` |
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