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mu observed = Sigma mu(i)X(i) where mu...

`mu` observed `= Sigma 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.

Text Solution

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Mole fraction of anti-form `=0.82`
Mole fraction of gauche form `=(1 - 0.82) = 0.18`
`mu_(obs) =1.0`
`mu_(obs) = mu_(1)X_(1) + mu_(2)X_(2)`
`1.0 = mu` (anti) `xx0.82 + mu` (gauche) `xx0.18`
`mu` (anti) = 0 (from Newman's projection, two Z vectors cancel and all H vectors cancel).
`1.0 = 0xx82 + mu` (gauche) `xx0.18`
`mu` (gauche) `=(1.0)/(0.18) =5.55 D`

In (i) anti form is a stable conformer.
In (ii), gauche form is a stable conformer due to the intramolecular H-bonding.
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