Question

A neutral water molecule in its vapour state has an electric dipole moment of 6.4×10^-30 cm how far are tge molecule centres of positive and negative charge

Answer 1

Answer:

Explanation:

Electric dipole moment of water molecule

P = 6.4 × 10^ -30 Cm

P = qd

Where d is the distance between the centre of positive and negative charge of the molecule

d = p/q = 6.4 × 10^-30 Cm/ 1.6 × 10^-19 C

= 4 × 10^ -11 m

Answer 2

Solution:

Dipole moment of a molecule, $p = q\times d$

where q = charge

d = distance between centres of positive and negative charge

In a neutral water molecule $(H_2O)$,

number of electrons = number of protons = 10

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 10 electron = $10\times1.6\times 10^{-19}C$

$p = q\times d\\\\d = \dfrac{p}{q} \\\\d = \dfrac{6.4\times 10^{-30}}{10\times 1.6\times 10^{-19}}$

$d = 4 \times 10^{-12}$ cm