Question

For the stable conformer of $Z-C{H}_2-C{H}_2-Z$
Given that ${\mu }_{obs}=\sum {\mu }_i{x}_i$
where ${\mu }_i$ is the dipole moment of the stable conformer and $x_i$ is the mole fraction of that conformer.
If ${\mu }_{solution}=1D$ and $x_{anti}=0.82$
Then value of ${\mu }_{gauche}$ is

Answer 1

Mole fraction of anti form (staggered) =0.82
Mole fraction of gauche form=1-0.82=0.18
Since eclipsed form is highly unstable, so $x_e\approx 0$
${\mu }_{solution}=\left({\mu }_s\right)\left(x_s\right)+\left({\mu }_g\right)\left(x_g\right)+\left({\mu }_e\right)\left(x_e\right)1=\left(0\right)\left(0.82\right)+\left({\mu }_g\right)\left(0.18\right)+\left({\mu }_e\right)\left(0\right)$
${\mu }_g=\frac{100}{18}=5.55$